Signals and systems Archives

Sampling Theorem

Sampling of signals is the fundamental operation in signal processing, a Continuous Time (CT) signal can be converted into a Discrete Time (DT) signal using Sampling process. Sampling is required since the advancement in both signals and systems which are digitized i.e, Digital systems operates only on digital signals only.

Sampling Theorem:-

A CT signal is first converted into DT signal by Sampling process. The sufficient number of samples must be taken so that the original signal is represented in it’s samples completely, and also the signal is represented from it’s samples, these two conditions representation and reconstruction depends on the sampling process ‘f_s‘ Hz.

Sampling theorem can be given into two parts

i. A band limited signal of finite energy, which has no frequency component higher than ‘f_m‘ Hz, is completely described by it’s sample values at uniform intervals less than (or) equal to 1/2f_m seconds apart.

i.e, $T_{s}\leq \frac{1}{2f_{m}}$ Seconds.

ii. A Band limited signal of finite energy, which has no frequency component higher than f_m Hz may be completely recovered from the knowledge of it’s samples if samples are taken at the rate of 2f_m samples/second.

i.e, $f_{s}\geq 2f_{m}$ Hz.

Statement:- A Continuous Time signal can be completely represented in it’s samples and recovered from it’s samples if the sampling frequency $f_{s}\geq 2f_{m}Hz.$

where $f_{s}$ is the sampling frequency.

$f_{m}$ is the highest frequency present in the original signal / Band width of the signal.

proof of Sampling theorem:-

Let us consider a CT signal x(t), which is a band limited to $f_{m}$ Hz as shown

To prove Sampling theorem, it should be shown a signal whose spectrum is band limited to f_m Hz can be reconstructed exactly without any error from it’s samples taken uniformly at a rate of $f_{s}> 2f_{m}$ Hz.

The circuit shows the sampler

Now sampling of x(t) at a rate of f_s may be achieved by multiplying x(t) with a train of impulses $\delta T_{s}(t)$ with a period ‘T_s‘ seconds.

The sampling signal is an ideal (or) instantaneous signal. This is also known as ideal (or) instantaneous sampling.

$g(t)=x(t)\delta T_{s}(t)$

As $\delta T_{s}(t)$ is a periodic impulse train it can be expressed in it’s Fourier Series expansion as follows

Exponential Fourier Series is

$\delta T_{s}(t) = \sum_{n=-\infty }^{\infty }F_{n}e^{jnw_{s}t}$

$F_{n}= \frac{1}{T_{s}}\int_{\frac{-T_{s}]}{2}}^{\frac{T_{s}}{2}}\delta T_{s}(t)e^{-jn\omega _{s}t}dt$

$F_{n}=\frac{1}{T_{s}}$

$F_{n}=f_{s}$

∴ Exponential Fourier Series is $\delta T_{s}(t)=\sum_{n=-\infty }^{\infty }f_{s}e^{jn\omega _{s}t}$

now the sampled signal $g(t) = x(t).\delta T_{s}(t)$

$g(t)=x(t)\sum_{n= -\infty }^{\infty }f_{s}e^{jn\omega _{s}t}$

$g(t)=\sum_{n= -\infty }^{\infty }f_{s}x(t)e^{jn\omega _{s}t}$

By finding Fourier Transform of g(t) is G(f)

$G(f)=\sum_{n= -\infty }^{\infty }f_{s}X(f-nf_{s})$

Now the frequency spectrum of the sampled signal G(f) is of the form

From G(f) spectrum the original spectrum of X(f) has been shifted to different center frequencies

i.e, when n=0 center frequency is 0.

n=1 center frequency is f_s

n=-1 center frequency is -f_s etc

Some important conclusions from frequency spectrum of sampled signal:-

The spectrum of sampled signal G(f)/G(w) will repeat periodically if $f_{s}> 2f_{m}$ without any overlapping.
G(f) is extending up to infinity and the Band width is infinity as well, out of G(f) , X(f) need to be recovered , which is band limited to f_m Hz.
X(f) is centered at f=0 and has f_m as the highest frequency, X(f) may be recovered by passing it through a Loe Pass filter with cutoff frequency approximately equals to f_m Hz.
to reconstruct x(t) from g(t) the condition that must be satisfied is $f_{s}\geq 2f_{m}$ .

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Band Pass Sampling

However when the given signal is a Band Pass signal then a different criterion must be used to sample the signal , the Band Pass signal x(t) whose maximum BW is ‘ $2f_{m}/W$ ‘ Hz can be completely represented and recovered from it’s samples if it is sampled at the minimum rate of greater than or equals to twice that of the BW.

then sampling rate $f_{s}\geq 2 X BW$

i.e, $f_{s}\geq 4f_{m} or f_{s}\geq 2W$

Any band pass signal in time-domain can be represented in it’s in-phase $x_{I}(t)$ and quadrature phase $x_{Q}(t)$ components as

$x(t) = x_{I}(t)cos 2\pi f_{c}t \pm x_{Q}(t)sin 2\pi f_{c}t$

after sampling the band pass signal, the signal after reconstruction is

$x(t) = \sum_{n=-\infty }^{\infty }sinc(2f_{m}t-\frac{n}{2})cos(2\pi f_{c}(t-\frac{n}{4f_{m}}))$

$T_{s} = \frac{1}{4f_{m}}$ , where BW of band pass signal is $2f_{m}$ Hz

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Reconstruction filter(Low Pass Filter)

Reconstruction filter (Low Pass Filter) Procedure to reconstruct actual signal from sampled signal:-

Low Pass Filter is used to recover original signal from it’s samples. This is also known as interpolation filter.

An LPF is that type of filter which passes only low frequencies up to cut-off frequency and rejects all other frequencies above cut-off frequency.

For an ideal LPF, there is a sharp change in the response at cut-off frequency as shown in the figure.

i.e, Amplitude response becomes suddenly zero at cut-off frequency which is not possible practically that means an ideal LPF is not physically realizable.

i.e, in place of an ideal LPF a practical filter is used.

In case of a practical filter, the amplitude response decreases slowly to zero (this is one of the reason why we choose $f_{s}>2f_{m}$ )

This means that there exists a transition band in case of practical Low Pass Filter in the reconstruction of original signal from its samples.

Signal Reconstruction (Interpolation function):-

The process of reconstructing a Continuous Time signal x(t) from it’s samples is known as interpolation.

Interpolation gives either approximate (or) exact reconstruction (or) recovery of CT signal.

One of the simplest interpolation procedures is known as zero-order hold.

Another procedure is linear interpolation. In linear interpolation the adjacent samples (or) sample points are connected by straight lines.

We may also use higher order interpolation formula for reconstructing the CT signal from its sample values.

If we use the above process (Higher order interpolation) the sample points are connected by higher order polynomials (or) other mathematical functions.

For a Band limited signal, if the sampling instants are sufficiently large then the signal may be reconstructed exactly by using a LPF.

In this case an exact interpolation can be carried out between sample points.

Mathematical analysis:-

A Band limited signal x(t) can be reconstructed completely from its samples, which has higher frequency component f_m Hz.

If we pass the sampled signal through a LPF having cut-off frequency of f_m Hz.

From sampling theorem

$g(t) = x(t).\delta _{T_{s}}(t)$ .

$g(t)=\frac{1}{T_{s}}\left \{ 1+2\cos \omega _{s}t+2\cos 2\omega _{s}t+2\cos 3\omega _{s}t+..... \right \}$ .

g(t) has a multiplication factor $\frac{1}{T_{s}}$ . To reconstruct x(t) (or) X(f) , the sampled signal must be passed through an ideal LPF of Band Width of $f_{m}$ Hz and gain $T_{s}$ .

$\left | H(\omega ) \right |=T_{s} \ for \ -\omega _{m}\leq \omega \leq \omega _{m}$ .

$h(t) = \frac{1}{2\pi } \int_{-\omega _{m}}^{\omega _{m}}T_{s}e^{j\omega t}\ d\omega$ .

$h(t) = 2f_{m}T_{s} \ sinc(2\pi f_{m}t)$ .

If sampling is done at Nyquist rate , then Nyquist interval is $T_{s} = \frac{1}{2f_{m}}$ .

therefore $h(t) = \ sinc(2\pi f_{m}t)$ .

h(t) = 0. at all Nyquist instants $t= \pm \frac{n}{2f_{m}}$ , when g(t) is applied at the input to this filter the output will be x(t) .

Each sample in g(t) results a sinc pulse having amplitude equal to the strength of sample. If we add all these sinc pulses that gives the original signal x(t) .

$g(t) = x(kT_{s})\delta (t-kT_{s})$ .

$x(t) =\sum_{k} x(kT_{s})\ h (t-kT_{s})$ .

$x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m} (t-kT_{s}))$ .

$x(t) =\sum_{k} x(kT_{s})\ sinc(2\pi f_{m}t-k\pi )$ .

This is known as interpolation formula

It is assumed that the signal x(t) is strictly band limited but in general an information signal may contain a wide range of frequencies and can not be strictly band limited this means that the maximum frequency in the signal can not be predictable.

then it is not possible to select suitable sampling frequency f_s .

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Properties of Z-Transforms (Bi-lateral)

1.Linearity Property:-

$x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}$

$x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}$

$a\ x_{1}[n]+b\ x_{2}[n]\leftrightarrow \ \ ?$ .

we know that Bi-lateral Z- Transform of a signal $x[n]$ is $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]+b\ x_{2}[n] \right \} Z^{-n}$ .

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]\ Z^{-n}+b\ x_{2}[n] \ Z^{-n}\right \}$

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a\ \sum_{n=-\infty }^{\infty } x_{1}[n]\ Z^{-n}+b \ \sum_{n=-\infty }^{\infty } x_{2}[n] \ Z^{-n}$ .

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a \ X_{1}(Z)+b \ X_{2}(Z)$ . $ROC: R_{1} \cap R_{2}$ .

2.Time-shifting Property:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC : R$

$x[n-k]\leftrightarrow \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ x[n-k] \right \} = \sum_{n=-\infty }^{\infty } x[n-k] \ Z^{-n}$ . Let $n-k=m \Rightarrow m= n+k$

Here n is a variable and k is a constant.

$Z\left \{ x[n-k] \right \} = \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-(m+k)}$ .

$Z\left \{ x[n-k] \right \} = Z^{-k} \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-m}$ .

$Z\left \{ x[n-k] \right \} = Z^{-k} X(Z)$

$x[n-k]\leftrightarrow \ Z^{-k} X(Z)\ , \ \ ROC:R$ .

from the above equation $x[n-k]$ forms Z Transform pair with $Z^{-k} X(Z)$ .

3. Scaling in-Z-domain property:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}$

$a^{n}x[n]\leftrightarrow \ \ \ ? ,$

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }a^{n}\ x[n] \ Z^{-n}$ .

$Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }\ x[n] \ (a^{-1}Z)^{-n}$ .

$Z\left \{ a^{n}\ x[n] \right \} = X(a^{-1}Z)$

$a^{n}x[n]\leftrightarrow \ X(\frac{Z}{a}), \ \ \ if \ a>0$ . $\ \ ROC \ of \ a^{n}x[n] \ is :\left | a \right |\ r_{1}< \left | Z \right | <\left | a \right |\ r_{2}$ .

4. Time-reversal property:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}$

$x[-n]\leftrightarrow \ \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ x[-n] \right \} = \sum_{n=-\infty }^{\infty } x[-n] \ Z^{-n}$ . Let ,

from the above equation $x[-n]$ forms Z Transform pair with $X(\frac{1}{Z})$ .

5. Differentiation in Z-domain:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}$

$n\ x[n]\leftrightarrow \ \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } x[n] \frac{d (Z^{-n})}{dZ}$ .

$\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } \ -n \ x[n] Z^{(-n-1)}$ .

$\frac{d(X(Z))}{dZ} = -\left \ Z^{-n}\right ] Z^{-1}$ .

$\frac{d(X(Z))}{dZ} = - Z^{-1} \ Z\left \{ n\ x[n] \right \}$ .

from the above equation $\frac{d(X(Z))}{dZ}$ forms Z-Transform pair with $n\ x[n]$ and the ROC is same as that of the original sequence x[n].

6. Convolution in Time-domain:-

$x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}$

$x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}$

$x_{1}[n]* x_{2}[n]\leftrightarrow \ \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty }(x_{1}[n] *x_{2}[n]) \ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }(x_{1}[k] x_{2}[n-k]) \sum_{n=-\infty }^{\infty }\ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \sum_{n=-\infty }^{\infty }x_{2}[n-k] \ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \ Z^{-k}\ X_{2}(Z)$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \left \ Z^{-k} \right ]\ X_{2}(Z)$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = X_{1}(Z) \ X_{2}(Z)$ .

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Derivation of Z and inverse Z-Transforms

Derivation of Z-Transform:-

If $x[n]$ is the given sequence then it’s Discrete Time Fourier Transform is $X(e^{j\omega })$ .

i.e, $x[n]\leftrightarrow X(e^{j\omega })$

$x[n]\ r^{-n}\leftrightarrow X(r\ e^{j\omega })$ .

DTFT of $x[n] = \sum_{n=-\infty }^{\infty } x[n] e^{-j\omega n}$ .

DTFT of $x[n]\ r^{-n} = \sum_{n=-\infty }^{\infty } x[n] \ r^{-n}e^{-j\omega n}$ .

$X(r\ e^{j\omega }) = \sum_{n=-\infty }^{\infty } x[n] \ \left ( r\ e^{j\omega } \right )^{-n}$ .

Let $Z=r\ e^{j\omega }$ a complex-variable.

$X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z ^{-n}$ . is the Z-Transform of a sequence $x[n]$ .

Inverse Z-Transform:-

The inverse DTFT of $X(e^{j\omega })$ is $x[n] = \frac{1}{2\pi } \int X(e^{j\omega }) \ e^{j\omega n} \ d\omega$ .

$x[n] \ r^{-n} = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ e^{j\omega n} \ d\omega$ .

$x[n] = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ (r\ e^{j\omega })^{n} \ d\omega$ .

$x[n] = \frac{1}{2\pi\ j Z } \int X(Z) \ Z^{n} \ dz$ . Let $r\ e^{j\omega } = Z \Rightarrow \j \ r \ e^{j\omega }\ d\omega = dZ$ and $\ j \ Z \ d\omega = dZ \Rightarrow d\omega =\frac{dz}{j \ Z}$ .

$x[n] = \frac{1}{2\pi\ j } \int X(Z) \ Z^{n-1} \ dz$ – represents the inverse Z-Transform.

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Initial and Final Value Theorems of Laplace Transforms

Initial-value Theorem:-

Use:- to find out the initial value of a signal x(t) without using inverse Laplace Transform.

$x(0^{-})= \lim_{t\rightarrow 0^{-}}x(t)=\lim_{S\rightarrow \infty }s\ X(S)$ .

Proof:-

we know that $L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-})$ .

$L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt$ .

i.e, $\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-})$ .

$\lim_{s\rightarrow \infty }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-})$ .

$0= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-})$ .

$\lim_{s\rightarrow \infty }\ S\ X(S)= \lim_{s\rightarrow \infty }\ x(0^{-})$ .

$\ x(0^{-}) = \lim_{s\rightarrow \infty }\ S\ X(S)$ .

Hence proved.

final-value Theorem:-

Use:- to find out the final value of a signal x(t) without using inverse Laplace Transform.

$x(\infty )= \lim_{t\rightarrow \infty }x(t)=\lim_{S\rightarrow 0 }s\ X(S)$ .

Proof:-

we know that $L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-})$ .

$L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt$ .

i.e, $\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-})$ .

$\lim_{s\rightarrow 0 }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-})$ .

$x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-})$ .

$x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- x(0^{-})$

$x(\infty )= \lim_{s\rightarrow 0 }\ S\ X(S)$ .

Hence proved.

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Fourier Series and it’s applications

The starting point of Fourier Series is the development of representation of signals as linear combination (sum of) of a set of basic signals.

$f(t)\approx C_{1}x_{1}(t)+C_{2}x_{2}(t)+.......+C_{n}x_{n}(t)+....$

The alternative representation if a set of complex exponentials are used,

$f(t)\approx C_{1}e^{j\omega _{o}t}+C_{2}e^{2j\omega _{o}t}+.......+C_{n}e^{jn\omega _{o}t}+....$

The resulting representations are known as Fourier Series in Continuous-Time . Here we focus on representation of Continuous-Time and Discrete-Time periodic signals in terms of basic signals as Fourier Series and extend the analysis to the Fourier Transform representation of broad classes of aperiodic, finite energy signals.

These Fourier Series & Fourier Transform representations are most powerful tools used

In the analyzation of signals and LTI systems.
Designing of Signals & Systems.
Gives insight to S&S.

The development of Fourier series analysis has a long history involving a great many individuals and the investigation of many different physical phenomena.

The concept of using “Trigonometric Sums”, that is sum of harmonically related sines and cosines (or) periodic complex exponentials are used to predict astronomical events.

Similarly, if we consider the vertical deflection $f(t,x)$ of the string at time t and at a distance x along the string then for any fixed instant of time, the normal modes are harmonically related sinusoidal functions of x.

The scientist Fourier’s work, which motivated him physically was the phenomenon of heat propagation and diffusion. So he found that the temperature distribution through a body can be represented by using harmonically related sinusoidal signals.

In addition to this he said that any periodic signal could be represented by such a series.

Fourier obtained a representation for aperiodic (or) non-periodic signals not as weighted sum of harmonically related sinusoidals but as weighted integrals of Sinusoids that are not harmonically related, which is known as Fourier Integral (or) Fourier Transform.

In mathematics, we use the analysis of Fourier Series and Integrals in

The theory of Integration.
Point-set topology.
and in the eigen function expansions.

In addition to the original studies of vibration and heat diffusion, there are numerous other problems in science and Engineering in which sinusoidal signals arise naturally, and therefore Fourier Series and Fourier T/F’s plays an important role.

for example, Sine signals arise naturally in describing the motion of the planets and the periodic behavior of the earth’s climate.

A.C current sources generate sinusoidal signals as voltages and currents. As we will see the tools of Fourier analysis enable us to analyze the response of an LTI system such as a circuit to such Sine inputs.

Waves in the ocean consists of the linear combination of sine waves with different spatial periods (or) wave lengths.

Signals transmitted by radio and T.V stations are sinusoidal in nature as well.

The problems of mathematical physics focus on phenomena in Continuous Time, the tools of Fourier analysis for DT signals and systems have their own distinct historical roots and equally rich set of applications.

In particular, DT concepts and methods are fundamental to the discipline of numerical analysis , formulas for the processing of discrete sets of data points to produce numerical approximations for interpolation and differentiation were being investigated.

FFT known as Fast Fourier Transform algorithm was developed, which suited perfectly for efficient digital implementation and it reduced the time required to compute transform by orders of magnitude (which utilizes the DTFS and DTFT practically).

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aliasing effect in Sampling

Effect of under sampling (aliasing effect):-

When a CT band limited signal is sampled at $f_{s} < 2f_{m}$ , then the successive cycles of the spectrum of the sampled signal overlap with each other as shown below

Some aliasing is produced in the signal this is due to under sampling.

aliasing is the phenomenon in which a high frequency component in the frequency spectrum of the signal takes as a low frequency component in the spectrum of the sampled signal.

Because of aliasing it is not possible to reconstruct x(t) from g(t) by low pass filtering.

The spectral components in the overlapping regions and hence the signal is distorted.

Since any information signal contains a large no.of frequencies so the decision of sampling frequency is always become a problem.

A signal is first passed through LPF before sampling.

i.e, it is band limited by this LPF which is known as pre-alias filter.

To avoid aliasing

Pre-alias filter must be used to limit the band width of the signal to $f_{m}$ Hz.
Sampling frequency must be

Pre-alias filter means before sampling is passed through a LPF to make a perfect band limited signal.

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Fourier Series and it’s applications

The starting point of Fourier Series is the development of representation of signals as linear combination (sum of) of a set of basic signals.

$f(t)\approx C_{1}x_{1}(t)+C_{2}x_{2}(t)+.......+C_{n}x_{n}(t)+....$

The alternative representation if a set of complex exponentials are used,

$f(t)\approx C_{1}e^{j\omega _{o}t}+C_{2}e^{2j\omega _{o}t}+.......+C_{n}e^{jn\omega _{o}t}+....$

These Fourier Series & Fourier Transform representations are most powerful tools used

In the analyzation of signals and LTI systems.
Designing of Signals & Systems.
Gives insight to S&S.

The development of Fourier series analysis has a long history involving a great many individuals and the investigation of many different physical phenomena.

The concept of using “Trigonometric Sums”, that is sum of harmonically related sines and cosines (or) periodic complex exponentials are used to predict astronomical events.

In addition to this he said that any periodic signal could be represented by such a series.

In mathematics, we use the analysis of Fourier Series and Integrals in

The theory of Integration.
Point-set topology.
and in the eigen function expansions.

for example, Sine signals arise naturally in describing the motion of the planets and the periodic behavior of the earth’s climate.

Waves in the ocean consists of the linear combination of sine waves with different spatial periods (or) wave lengths.

Signals transmitted by radio and T.V stations are sinusoidal in nature as well.

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Relation between Laplace and Fourier Transform

The Fourier transform of a signal x(t) is given as

$X(j\omega ) = \int_{-\infty }^{\infty } x(t) e^{-j\omega t}dt----EQN(I)$

Fourier Transform exists only if $\int_{-\infty }^{\infty } \left | x(t) \right |dt< \infty$

we know that $s=\sigma + j\omega$

$X(S) = \int_{-\infty }^{\infty } x(t) e^{-s t}dt$

$X(S) = \int_{-\infty }^{\infty } \left | x(t)e^{-\sigma t} \right | e^{-j\omega t}dt----EQN(II)$

if we compare Equations (I) and (II) both are equal when $\sigma =0$ .

i.e, $X(S) =X(j\omega)| \right |_{s=j\omega }$ .

This means that Laplace Transform is same as Fourier transform when $s=j\omega$ .

Fourier Transform is nothing but the special case of Laplace transform where $s=j\omega$ indicates the imaginary axis in complex-s-plane.

Thus Laplace transform is basically Fourier Transform on imaginary axis in the s-plane.

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Region of Convergence (ROC)

The range of values of the complex variable s for which Laplace Transform $X(S)=\int_{-\infty }^{\infty }x(t)e^{-st}\ dt$ converges is called the Region of Convergence (ROC).

i.e, The region of Convergence (or) existence of signal’s Laplace transform X(S) is the set of values of s for which the integral defining the direct L T/F X(S) converges.

The ROC is required for evaluating the inverse L T/F of x(t) from X(S).

i.e, the operation of finding the inverse T/F requires an integration in the complex plane.

i.e, $x(t)= \frac{1}{2\pi j}\int_{\sigma -j\infty }^{\sigma +j\infty }X(S) e^{St} \ ds$ .

The path of integration is along S-plane $S = \sigma +j\omega$ that is along $\sigma +j\omega$ with $\omega$ varying from $-\infty \ to \ \infty$ and moreover , the path of integration must lie in the ROC for X(S).

for example the signal $e^{-at}u(t)$ , this is possible if $\sigma >-a$ so the path of integration is shown in the figure

Thus to obtain $x(t) = e^{-at}u(t)$ from $X(S) = \frac{1}{s+a}$ , the integration is performed through this path for the function $\frac{1}{s+a}$ . such integration in the complex plane requires a back ground in the theory of functions of complex variables.

so we can avoid this integration by compiling a Table of L T/F’s . so for inverse L T/F’s we use this table instead of performing complex integration.

specific constraints on the ROC are closely associated with time-domain properties of x(t).

Properties of ROC/ constraints (or) Limitations:-

1.The ROC of X(S) consists of strips parallel to the $j\omega$ axis in the S-plane.

i.e, The ROC of X(S) consists of the values of s for which Fourier T/F of $x(t)e^{-\sigma t}$ converges this is possible if $x(t)e^{-\sigma t}$ is fully integrable thus the condition depends only on $\sigma$ . Hence ROC is the strips (bands) which is only in terms of values of $\sigma$ .

3. For Rational Laplace T/F’s , the ROC does not contain any poles. This is because X(S) is finite at poles and the integral can not be converge at this point.

4. If x(t) is of finite duration and absolutely integrable, then the ROC is the entire S-plane.

5. If x(t) is right-sided and if the line $Re\left \{ s \right \} =\sigma _{o}$ is in the ROC, then all values of s for which $Re\left \{ s \right \} > \sigma _{o}$ will also be in the ROC.

i.e, if the signal is $x(t) = e^{-at}u(t)$ right-sided $[0 \ to \ \infty ]$ then $X(S) = \frac{1}{s+a}$ for ROC : $Re\left \{ s \right \} > -a$ .

6. If x(t) is left-sided and if the line $Re\left \{ s \right \} =\sigma _{o}$ is in the ROC, then all values of s for which $Re\left \{ s \right \} < \sigma _{o}$ will also be in the ROC.

7. If x(t) is two-sided and if the line $Re\left \{ s \right \} =\sigma _{o}$ is in the ROC, then the ROC consists of a strip in the s-plane that includes the line $Re\left \{ s \right \} = \sigma _{o}$ .

for the both sided signal , the ROC lies in the region $\sigma _{1} < Re\left \{ s \right \}<\sigma _{2}$ . This ROC is the strip parallel to $j\omega$ axis in the s-plane.

8. If the L T/F X(S) of x(t) is rational, then it’s ROC is bounded by poles (or) extends to infinity in addition no poles of X(s) are contained in the ROC.

If the function has two poles , then ROC will be area between these two poles for two sided signal, if for single sided signal the area extends from one pole to infinity.

But is does not include any pole.

9. If the L T/F X(S) of x(t) is rational, then if x(t) is right-sided. The ROC is the region in the s-plane to the right of the right most pole and if x(t) is left-sided, the ROC is the region in the s-plane to the left of the left most pole.

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Analogy between Vectors and Signals

we have already defined the signal as any ordinary function of time. To understand more about signal we consider it as a problem. A problem is better understood (or) better remembered if it can be associated with some familiar phenomenon.

we always search for analogies while studying a new problem.

i.e, In the study of abstract problems analogies are very helpful. Particularly if the problem can be shown to be analogous to some concrete phenomenon.

It is then easier to gain some insight into the new problem from the knowledge of the analogous phenomenon.

There is a perfect analogy that exists between vectors and signals which leads to a better understanding of signal analysis. we shall now briefly review the properties of vectors.

Vectors:-

A vector is specified by magnitude and direction $\overrightarrow{A}$ .

Let us consider two vectors $\overrightarrow{V_{1}}$ and $\overrightarrow{V_{2}}$ . It is possible to find out the component of one vector along the other vector.

In order to find out the component of vector $\overrightarrow{V_{1}}$ along $\overrightarrow{V_{2}}$ . Let us assume it as $C_{12}V_{2}$ , which is only the magnitude.

how do we represent physically the component of one vector $\overrightarrow{V_{1}}$ along $\overrightarrow{V_{2}}$ ? This is possible by finding the projection of one vector on to the other.

i.e, by drawing a perpendicular from $\overrightarrow{V_{1}}$ to $\overrightarrow{V_{2}}$

$\overrightarrow{V_{1}} = C_{12}\overrightarrow{V_{2}} + \overrightarrow{V_{e}}$ .

There exists two other possibilities.

but these are not suitable. $\because$ the error vectors are more in these cases.

$\overrightarrow{V_{1}}.\overrightarrow{V_{2}}=V_{1}V_{2}\cos \theta$ .

If $\theta$ is the angle between two vectors $\overrightarrow{V_{1}}$ and $\overrightarrow{V_{2}}$ , the component of $\overrightarrow{V_{1}}$ along $\overrightarrow{V_{2}}$ is

$\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left | V_{2} \right |}=V_{1}\cos \theta-----EQN(1)$ .

The component of $\overrightarrow{V_{1}}$ along $\overrightarrow{V_{2}}$ is $C_{12}V_{2}----EQN(2)$ .

$\therefore (1) = (2)$ .

$\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{\left | V_{2} \right |} = C_{12}V_{2}$ .

$C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}}$ .

If two vectors are orthogonal $\overrightarrow{V_{1}}.\overrightarrow{V_{2}} =0$ .

i.e, $C_{12} =0$ .

Signals:-

The concept of vector comparison & orthogonality can be extended to signals.

i.e, a signal is nothing but a single-valued function of independent variable. Assume two signals $f_{1}(t)$ and $f_{2}(t)$ , now to approximate $f_{1}(t)$ in terms of $f_{2}(t)$ over $t_{1}< t<t_{2}$ .

$f_{1}(t) \approx C_{12}f_{2}(t)$ .

$\therefore f_{1}(t) \approx C_{12}f_{2}(t)+f_{e}(t)$ .

$f_{e}(t) = f_{1}(t)-C_{12}f_{2}(t)$ .

Now, we choose in order to achieve the best approximation.

i.e, which keeps the error as minimum as possible.

One possible way for minimizing error $f_{e}(t)$ is to choose minimize the average value of $f_{e}(t)$ .

i.e, as $\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}dt$ .

But the process of averaging gives a false indication.

i.e, for example while approximating a function $\sin t$ with a null function $f(t)=0$ is

$f_{1}(t) =C_{12}f_{2}(t)$ .

$\sin t =0. \ \ 0\leq t\leq 2\pi$ .

indicates that $\sin t =0$ during $0$ to $2\pi$ without any error

i.e, $f_{e}(t) = f_{1}(t)-C_{12}f_{2}(t)$ .

$f_{e}(t) = \sin t$ .

Average value of error is $= \frac{1}{2 \pi } \int_{0}^{2\pi } \sin t \ dt =0$ .

This seems to be error is zero but actually there exists some error.

To avoid this false indication, we choose to minimize the average of the square of the error

i.e, Mean Square Error $\epsilon = \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}f_{e}^{2}(t) \ dt$ .

$\epsilon = \frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}(f_{1}(t)-C_{12}f_{2}(t))^{2} \ dt$ .

To find value which keeps error minimum $\frac{d\epsilon }{dC_{12}}=0$ .

$C_{12} = \frac{\int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt}{\int_{t_{1}}^{t_{2}}f_{2}^{2}(t) \ dt}$ .

$C_{12}$ Which is similar to $C_{12}=\frac{\overrightarrow{V_{1}}.\overrightarrow{V_{2}}}{V_{2}^{2}}$ where $\int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt$ denotes the inner product between two Real signals

$\therefore$ For the orthogonality of two signals $C_{12} =0$

$\Rightarrow \ \int_{t_{1}}^{t_{2}}f_{1}(t)f_{2}(t) \ dt =0$ .

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Properties of Laplace Transforms(Bi-lateral)

1.Linearity Property:-

$x_{1}(t)\leftrightarrow X_{1}(S) \ \ \ ROC : R_{1}$

$x_{2}(t)\leftrightarrow X_{2}(S) \ \ \ ROC : R_{2}$

$a\ x_{1}(t)+b\ x_{2}(t)\leftrightarrow \ \ ?$ .

we know that Laplace Transform of a signal $x(t)$ is $X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt$ .

$L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)+b\ x_{2}(t) \right \} \ e^{-St}\ dt$ .

$L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)\ e^{-St}\ dt+b\ x_{2}(t) \ e^{-St}\ dt\right \}$

$L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a \int_{-\infty }^{\infty }\ x_{1}(t)\ e^{-St}\ dt+b \int_{-\infty }^{\infty }\ x_{2}(t) \ e^{-St}\ dt$ .

$L\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a X_{1}(S) +b X_{2}(S)$ . $ROC: R_{1} \cap R_{2}$ .

2.Time-shifting Property:-

$x(t)\leftrightarrow X(S) \ \ \ ROC : R$

$x(t-t_{o})\leftrightarrow \ ?$ .

we know that $X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt$ .

$L\left \{ x(t-t_{o}) \right \} = \int_{-\infty }^{\infty }\ x(t-t_{o}) \ e^{-St}\ dt$ . Let $t-t_{o}=\lambda \Rightarrow dt= d\lambda$

$t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty$

$L\left \{ x(t-t_{o}) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-S(\lambda +t_{o})}\ d\lambda$ .

$L\left \{ x(t-t_{o}) \right \} =e^{-St_{o}} \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-S\lambda }\ d\lambda$ .

$x(t-t_{o})\leftrightarrow \ e^{-St_{o}}\ X(S) \ , \ \ ROC:R$ .

from the above equation $x(t-t_{o})$ forms a Laplace Transform pair with $e^{-St_{o}} \ X(S)$ .

3.Frequency-shifting Property:-

$x(t)\leftrightarrow X(S) \ \ \ ROC : R$

$?\ \leftrightarrow \ X(S-S_{o})$ .

we know that $X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt$ .

$L\left \{ e^{S_{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ e^{S_{o}t}\ x(t) \ e^{-St}\ dt$ .

$L\left \{ e^{S_{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ \ x(t) \ e^{-(S-S_{o})t}\ dt$ .

$e^{S_{o}t}x(t)\leftrightarrow \ X(S-S_{o}) \ , \ \ ROC:R$ .

from the above equation $e^{S_{o}t}x(t)$ forms a Laplace Transform pair with $X(S-S_{o})$ .

4. Differentiation in time-domain:-

$x(t)\leftrightarrow X(S) \ \ \ ROC : R$

$\frac{dx(t)}{dt}\leftrightarrow \ ?$ .

we know that inverse Laplace Transform $x(t) =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ e^{St} \ dS$ .

$\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ \frac{d(e^{St})}{dt} \ dS$ .

$\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty }\ X(S) \ S \ e^{St} \ dS$ .

$\frac{dx(t)}{dt} =\frac{1}{2\pi \ j} \int_{\sigma - j\infty }^{\sigma +j\infty } (\ S\ X(S)) \ e^{St} \ dS$ .

$\frac{dx(t)}{dt}\leftrightarrow \ S\ X(S)$ .

from the above equation $\frac{dx(t)}{dt}$ forms a Laplace Transform pair with $S\ X(S)$

Similarly $\frac{d^{n}x(t)}{dt^{n}}\leftrightarrow \ S^{n}\ X(S)$ .

5.Differentiation in S-domain:-

$x(t)\leftrightarrow X(S) \ \ \ ROC : R$

$?\leftrightarrow \frac{dX(S)}{dS}$ .

we know that $X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt$ .

$\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ x(t) \ \frac{de^{-St} }{dS}\ dt$ .

$\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ (-t)\ dt$ .

$\frac{dX(S)}{dS} = \int_{-\infty }^{\infty }\ (-t \ x(t)) \ e^{-St} \ dt$ .

$\frac{dX(S)}{dS}\leftrightarrow \ -t\ x(t) \ \ \ ROC:R$ .

from the above equation $\frac{dX(S)}{dS}$ forms a Laplace Transform pair with $-t\ x(t)$ .

6. Time-reversal property:-

$x(t)\leftrightarrow X(S) \ \ \ ROC : R$

$x(-t)\leftrightarrow \ \ ?$ .

we know that $X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt$ .

$L\left \{ x(-t) \right \} = \int_{-\infty }^{\infty }\ x(-t) \ e^{-St} \ dt$ . Let $-t=\ \lambda$ , $-dt=\ d\lambda$ , $t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty$ .

$L\left \{ x(-t) \right \} = \int_{\lambda =\infty }^{-\infty }\ x(\lambda ) \ e^{S\lambda } \ (-d\lambda )$ .

$L\left \{ x(-t) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-(-S) \lambda } \ d\lambda$ .

$x(-t)\leftrightarrow \ X(-S)$ .

from the above equation $x(-t)$ forms a Laplace Transform pair with $X(-S)$ .

7. Time-Scaling property:-

$x(t)\leftrightarrow X(S) \ \ \ ROC : R$

$x(at)\leftrightarrow \ \ ?$ .

we know that $X(S) = \int_{-\infty }^{\infty }\ x(t) \ e^{-St} \ dt$ .

$L\left \{ x(at) \right \} = \int_{-\infty }^{\infty }\ x(at) \ e^{-St} \ dt$ . Let $at=\ \lambda$ , $dt=\ \frac{d\lambda}{a}$ , $t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ -\infty \ to \ \infty$ .

$L\left \{ x(at) \right \} = \frac{1}{a}\int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{(\frac{-S}{a})\lambda } \ d\lambda$ .

$x(at)\leftrightarrow \frac{1}{a} \ X(\frac{S}{a}), \ \ \ if \ a>0$ .

$x(-at)\leftrightarrow \frac{1}{a} \ X(\frac{-S}{a}), \ \ \ if \ a<0 \ and \ (a\neq -1)$ .

8. Convolution in Time-domain:-

$x_{1}(t)\leftrightarrow X_{1}(S) \ \ \ ROC : R_{1}$

$x_{2}(t)\leftrightarrow X_{2}(S) \ \ \ ROC : R_{2}$

$x_{1}(t) * x_{2}(t)\leftrightarrow \ \ ?$ .

we know that $X(S) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-St} \ dt$ .

$L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \left \{ x_{1}(t) * x_{2}(t) \right \}\ e^{-St} \ dt$ .

$L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ x_{2}(t-\tau ) \right \}\ e^{-St} \ dt \ d\tau$ .

$L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ \int_{t=-\infty }^{\infty } x_{2}(t-\tau )\ e^{-St} \ dt \right \} \ d\tau$ .

$L\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ e^{-S\tau } X_{2}(S) \right \} \ d\tau$ .

$L\left \{ x_{1}(t) * x_{2}(t) \right \} = \left \{ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau ) e^{-S\tau } \ d\tau \right \} \ X_{2}(S)$ .

$L\left \{ x_{1}(t) * x_{2}(t) \right \} = \ X_{1}(S) \ X_{2}(S)$

$x_{1}(t) * x_{2}(t)\leftrightarrow \ X_{1}(S) \ X_{2}(S), ROC : R_{1} \cap R_{2}$ .

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Initial-value & Final -value Theorems:-

Initial-value Theorem:-

Use:- to find out the initial value of a signal x(t) without using inverse Laplace Transform.

$x(0^{-})= \lim_{t\rightarrow 0^{-}}x(t)=\lim_{S\rightarrow \infty }s\ X(S)$ .

Proof:-

we know that $L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-})$ .

$L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt$ .

i.e, $\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-})$ .

$\lim_{s\rightarrow \infty }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-})$ .

$0= \lim_{s\rightarrow \infty }\ S\ X(S)- \lim_{s\rightarrow \infty }\ x(0^{-})$ .

$\lim_{s\rightarrow \infty }\ S\ X(S)= \lim_{s\rightarrow \infty }\ x(0^{-})$ .

$\ x(0^{-}) = \lim_{s\rightarrow \infty }\ S\ X(S)$ .

Hence proved.

final-value Theorem:-

Use:- to find out the final value of a signal x(t) without using inverse Laplace Transform.

$x(\infty )= \lim_{t\rightarrow \infty }x(t)=\lim_{S\rightarrow 0 }s\ X(S)$ .

Proof:-

we know that $L\left \{ \frac{dx(t)}{dt} \right \}\leftrightarrow S\ X(S)-x(0^{-})$ .

$L\left \{ \frac{dx(t)}{dt} \right \}=\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt$ .

i.e, $\int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt=S\ X(S)-x(0^{-})$ .

$\lim_{s\rightarrow 0 }\ \int_{0^{-}}^{\infty } \frac{dx(t)}{dt}\ e^{-St}\ dt= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-})$ .

$x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- \lim_{s\rightarrow 0 }\ x(0^{-})$ .

$x(\infty )-x(0^{-})= \lim_{s\rightarrow 0 }\ S\ X(S)- x(0^{-})$

$x(\infty )= \lim_{s\rightarrow 0 }\ S\ X(S)$ .

Hence proved.

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Derivation of Z-Transform/ inverse Z- Transform

Derivation of Z-Transform:-

If $x[n]$ is the given sequence then it’s Discrete Time Fourier Transform is $X(e^{j\omega })$ .

i.e, $x[n]\leftrightarrow X(e^{j\omega })$

$x[n]\ r^{-n}\leftrightarrow X(r\ e^{j\omega })$ .

DTFT of $x[n] = \sum_{n=-\infty }^{\infty } x[n] e^{-j\omega n}$ .

DTFT of $x[n]\ r^{-n} = \sum_{n=-\infty }^{\infty } x[n] \ r^{-n}e^{-j\omega n}$ .

$X(r\ e^{j\omega }) = \sum_{n=-\infty }^{\infty } x[n] \ \left ( r\ e^{j\omega } \right )^{-n}$ .

Let $Z=r\ e^{j\omega }$ a complex-variable.

$X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z ^{-n}$ . is the Z-Transform of a sequence $x[n]$ .

Inverse Z-Transform:-

The inverse DTFT of $X(e^{j\omega })$ is $x[n] = \frac{1}{2\pi } \int X(e^{j\omega }) \ e^{j\omega n} \ d\omega$ .

$x[n] \ r^{-n} = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ e^{j\omega n} \ d\omega$ .

$x[n] = \frac{1}{2\pi } \int X(r\ e^{j\omega }) \ (r\ e^{j\omega })^{n} \ d\omega$ .

$x[n] = \frac{1}{2\pi\ j Z } \int X(Z) \ Z^{n} \ dz$ . Let $r\ e^{j\omega } = Z \Rightarrow \j \ r \ e^{j\omega }\ d\omega = dZ$ and $\ j \ Z \ d\omega = dZ \Rightarrow d\omega =\frac{dz}{j \ Z}$ .

$x[n] = \frac{1}{2\pi\ j } \int X(Z) \ Z^{n-1} \ dz$ – represents the inverse Z-Transform.

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properties of Fourier Transforms

1.Linearity Property:-

$x_{1}(t)\leftrightarrow X_{1}(j\omega )$

$x_{2}(t)\leftrightarrow X_{2}(j\omega)$

$a\ x_{1}(t)+b\ x_{2}(t)\leftrightarrow \ \ ?$ .

we know that Fourier Transform of a signal $x(t)$ is $X(j\omega) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt$ .

$F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)+b\ x_{2}(t) \right \} \ e^{-j\omega t}\ dt$ .

$F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \int_{-\infty }^{\infty }\ \left \{ a\ x_{1}(t)\ e^{-j\omega t}\ dt+b\ x_{2}(t) \ e^{-j\omega t}\ dt\right \}$

$F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ \ a \int_{-\infty }^{\infty }\ x_{1}(t)\ e^{-j\omega t}\ dt+b \int_{-\infty }^{\infty }\ x_{2}(t) \ e^{-j\omega t}\ dt$ .

$F\left \{a\ x_{1}(t)+b\ x_{2}(t)\right \} = \ a \ X_{1}(j\omega) +b \ X_{2}(j\omega)$ .

2.Time-shifting Property:-

$x(t)\leftrightarrow X(j\omega )$

$x(t-t_{o})\leftrightarrow \ ?$ .

we know that $X(j\omega) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt$ .

$L\left \{ x(t-t_{o}) \right \} = \int_{-\infty }^{\infty }\ x(t-t_{o}) \ e^{- j\omega t}\ dt$ . Let $t-t_{o}=\lambda \Rightarrow dt= d\lambda$

$t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ \infty \ to \ -\infty$

$L\left \{ x(t-t_{o}) \right \} = \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{- j\omega (\lambda +t_{o})}\ d\lambda$ .

$L\left \{ x(t-t_{o}) \right \} =e^{-j\omega t_{o}} \int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-j\omega \lambda }\ d\lambda$ .

$x(t-t_{o})\leftrightarrow \ e^{- j\omega t_{o}}\ X(j\omega)$ .

from the above equation $x(t-t_{o})$ forms Fourier Transform pair with $e^{- j\omega t_{o}} \ X(j\omega)$ .

3.Frequency-shifting Property:-

$x(t)\leftrightarrow X(\omega )$

$?\ \leftrightarrow \ X(\omega -\omega _{o})$ .

we know that $X(\omega ) \ or X(j\omega ) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt$ .

$L\left \{ e^{j\omega _{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ e^{j \omega _{o}t}\ x(t) \ e^{-j\omega t}\ dt$ .

$L\left \{ e^{j\omega _{o}t}\ x(t) \right \} = \int_{t=-\infty }^{\infty }\ \ x(t) \ e^{-(\omega -\omega _{o})t}\ dt$ .

$e^{j\omega _{o}t}x(t)\leftrightarrow \ X(\omega -\omega _{o})$ .

from the above equation $e^{j\omega _{o}t}\ x(t)$ forms Fourier Transform pair with $X(\omega -\omega _{o})$ .

4. Differentiation in time-domain:-

$x(t)\leftrightarrow X(\omega )$

$\frac{dx(t)}{dt}\leftrightarrow \ ?$ .

we know that inverse Fourier Transform $x(t) =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X( \omega ) \ e^{j\omega t} \ d\omega$ .

$\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X(\omega ) \ \frac{d(e^{j\omega t})}{dt} \ d\omega$ .

$\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ X(\omega ) \ j \omega \ e^{j\omega t} \ d\omega$ .

$\frac{dx(t)}{dt} =\frac{1}{2\pi } \int_{\omega =\infty }^{\infty }\ (\ j \omega \ X(\omega )) \ e^{j\omega t} \ d\omega$ .

$\frac{dx(t)}{dt}\leftrightarrow \ j \omega \ X(\omega )$ .

from the above equation $\frac{dx(t)}{dt}$ forms Fourier Transform pair with $\ j \omega \ X(\omega )$

Similarly $\frac{d^{n}x(t)}{dt^{n}}\leftrightarrow \ \ (j \omega) ^{n}\ X(\omega )$ .

5.Differentiation in w-domain:-

$x(t)\leftrightarrow X(\omega )$

$?\leftrightarrow \frac{dX(\omega )}{d\omega }$ .

we know that $X(\omega ) = \int_{t =-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt$ .

$\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ x(t) \ \frac{d(e^{-j\omega t}) }{d\omega } \ dt$ .

$\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ (-j t)\ dt$ .

$\frac{dX(\omega )}{d\omega } = \int_{t=-\infty }^{\infty }\ (-jt \ x(t)) \ e^{-j\omega t} \ dt$ .

$\frac{dX(\omega )}{d\omega }\leftrightarrow \ -jt\ x(t)$ .

from the above equation $\frac{dX(\omega )}{d\omega }$ forms Fourier Transform pair with $-jt\ x(t)$ .

6. Conjugation property:-

$x(t)\leftrightarrow X(\omega )$

$x^{*}(t)\leftrightarrow \ \ ?$ .

we know that $X(\omega ) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt$ .

$F\left \{ x^{*}(t) \right \} = \int_{-\infty }^{\infty }\ x^{*}(t) \ e^{-j\omega t} \ dt$ .

$F\left \{ x^{*}(t) \right \} = \int_{t =-\infty }^{\infty }( x(t ) \ e^{j \omega t} \ dt )^{*}$ .

$x^{*}(t)\leftrightarrow \ X^{*}(-\omega )$ .

from the above equation $x^{*}(t)$ forms Fourier Transform pair with $X^{*}(-\omega )$ .

7. Time-Scaling property:-

$x(t)\leftrightarrow X(\omega )$

$x(at)\leftrightarrow \ \ ?$ .

we know that $X(\omega ) = \int_{-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt$ .

$F\left \{ x(at) \right \} = \int_{-\infty }^{\infty }\ x(at) \ e^{-j\omega t} \ dt$ . Let $at=\ \lambda$ , $dt=\ \frac{d\lambda}{a}$ , $t \ limits \ : \ -\infty \ to \ \infty, \ \ \ \lambda \ \ limits \ : \ -\infty \ to \ \infty$ .

$F\left \{ x(at) \right \} = \frac{1}{a}\int_{\lambda =-\infty }^{\infty }\ x(\lambda ) \ e^{-j(\frac{\omega }{a})\lambda } \ d\lambda$ .

$x(at)\leftrightarrow \frac{1}{a} \ X(\frac{\omega }{a}), \ \ \ if \ a>0$ .

$x(-at)\leftrightarrow \frac{1}{a} \ X(\frac{-\omega }{a}), \ \ \ if \ a<0 \ and \ (a\neq -1)$ .

8. Convolution in Time-domain:-

$x_{1}(t)\leftrightarrow X_{1}(\omega )$

$x_{2}(t)\leftrightarrow X_{2}(\omega )$

$x_{1}(t) * x_{2}(t)\leftrightarrow \ \ ?$ .

we know that $X(\omega ) = \int_{t=-\infty }^{\infty }\ x(t) \ e^{-j\omega t} \ dt$ .

$F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \left \{ x_{1}(t) * x_{2}(t) \right \}\ e^{-j\omega t} \ dt$ .

$F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{t=-\infty }^{\infty }\ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ x_{2}(t-\tau ) \right \}\ e^{-j\omega t} \ dt \ d\tau$ .

$F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ \int_{t=-\infty }^{\infty } x_{2}(t-\tau )\ e^{-j\omega t} \ dt \right \} \ d\tau$ .

$F\left \{ x_{1}(t) * x_{2}(t) \right \} = \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau )\left \{ e^{-j\omega \tau } X_{2}(\omega ) \right \} \ d\tau$ .

$F\left \{ x_{1}(t) * x_{2}(t) \right \} = \left \{ \int_{\tau =-\infty }^{\infty }\ x_{1}(\tau ) e^{-j\omega \tau } \ d\tau \right \} \ X_{2}(\omega )$ .

$F\left \{ x_{1}(t) * x_{2}(t) \right \} = \ X_{1}(\omega ) \ X_{2}(\omega )$

$x_{1}(t) * x_{2}(t)\leftrightarrow \ X_{1}(\omega ) \ X_{2}(\omega )$ .

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Properties of Z-transforms

1.Linearity Property:-

$x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}$

$x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}$

$a\ x_{1}[n]+b\ x_{2}[n]\leftrightarrow \ \ ?$ .

we know that Bi-lateral Z- Transform of a signal $x[n]$ is $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]+b\ x_{2}[n] \right \} Z^{-n}$ .

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty } \ \ \left \{ a\ x_{1}[n]\ Z^{-n}+b\ x_{2}[n] \ Z^{-n}\right \}$

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a\ \sum_{n=-\infty }^{\infty } x_{1}[n]\ Z^{-n}+b \ \sum_{n=-\infty }^{\infty } x_{2}[n] \ Z^{-n}$ .

$Z\left \{a\ x_{1}[n]+b\ x_{2}[n]\right \} = \ a \ X_{1}(Z)+b \ X_{2}(Z)$ . $ROC: R_{1} \cap R_{2}$ .

2.Time-shifting Property:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC : R$

$x[n-k]\leftrightarrow \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ x[n-k] \right \} = \sum_{n=-\infty }^{\infty } x[n-k] \ Z^{-n}$ . Let $n-k=m \Rightarrow m= n+k$

Here n is a variable and k is a constant.

$Z\left \{ x[n-k] \right \} = \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-(m+k)}$ .

$Z\left \{ x[n-k] \right \} = Z^{-k} \sum_{m\ =-\infty }^{\infty } x[m] \ Z^{-m}$ .

$Z\left \{ x[n-k] \right \} = Z^{-k} X(Z)$

$x[n-k]\leftrightarrow \ Z^{-k} X(Z)\ , \ \ ROC:R$ .

from the above equation $x[n-k]$ forms Z Transform pair with $Z^{-k} X(Z)$ .

3. Scaling in-Z-domain property:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}$

$a^{n}x[n]\leftrightarrow \ \ \ ? ,$

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }a^{n}\ x[n] \ Z^{-n}$ .

$Z\left \{ a^{n}\ x[n] \right \} = \sum_{n=-\infty }^{\infty }\ x[n] \ (a^{-1}Z)^{-n}$ .

$Z\left \{ a^{n}\ x[n] \right \} = X(a^{-1}Z)$

$a^{n}x[n]\leftrightarrow \ X(\frac{Z}{a}), \ \ \ if \ a>0$ . $\ \ ROC \ of \ a^{n}x[n] \ is :\left | a \right |\ r_{1}< \left | Z \right | <\left | a \right |\ r_{2}$ .

4. Time-reversal property:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}$

$x[-n]\leftrightarrow \ \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ x[-n] \right \} = \sum_{n=-\infty }^{\infty } x[-n] \ Z^{-n}$ . Let $-n=\ m$ ,

$Z\left \{ x[-n] \right \} = \sum_{m=-\infty }^{\infty } x[m] \ (Z^{-1})^{-m}\ \ \ ROC :r_{1}< \left | Z^{-1} \right | <r_{2}$ .

$x[-n]\leftrightarrow \ X(Z^{-1})\ \ \ ROC :\frac{1}{\left | r_{2} \right |}< \left | Z \right | <\frac{1}{\left | r_{1} \right |}$ .

from the above equation $x[-n]$ forms Z Transform pair with $X(\frac{1}{Z})$ .

5. Differentiation in Z-domain:-

$x[n]\leftrightarrow X(Z) \ \ \ ROC :r_{1}< \left | Z \right | <r_{2}-R_{1}$

$n\ x[n]\leftrightarrow \ \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } x[n] \frac{d (Z^{-n})}{dZ}$ .

$\frac{d(X(Z))}{dZ} = \sum_{n=-\infty }^{\infty } \ -n \ x[n] Z^{(-n-1)}$ .

$\frac{d(X(Z))}{dZ} = -\left \ Z^{-n}\right ] Z^{-1}$ .

$\frac{d(X(Z))}{dZ} = - Z^{-1} \ Z\left \{ n\ x[n] \right \}$ .

from the above equation $\frac{d(X(Z))}{dZ}$ forms Z-Transform pair with $n\ x[n]$ and the ROC is same as that of the original sequence x[n].

6. Convolution in Time-domain:-

$x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}$

$x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}$

$x_{1}[n]* x_{2}[n]\leftrightarrow \ \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{n=-\infty }^{\infty }(x_{1}[n] *x_{2}[n]) \ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }(x_{1}[k] x_{2}[n-k]) \sum_{n=-\infty }^{\infty }\ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \sum_{n=-\infty }^{\infty }x_{2}[n-k] \ Z^{-n}$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \sum_{k=-\infty }^{\infty }x_{1}[k] \ Z^{-k}\ X_{2}(Z)$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = \left \ Z^{-k} \right ]\ X_{2}(Z)$ .

$Z\left \{ x_{1}[n] *x_{2}[n]\right \} = X_{1}(Z) \ X_{2}(Z)$ .

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Complex Convolution Theorem :-

Complex Convolution Theorem :-

Let two signals $x_{1}[n] , \ x_{2}[n]$ which are complex signals then the product of these two signals be $x_{3}[n] = x_{1}[n]\ x_{2}[n]$ .

$x_{1}[n]\leftrightarrow X_{1}(Z) \ \ \ ROC :a_{1}< \left | Z \right | <b_{1}-R_{1}$

$x_{2}[n]\leftrightarrow X_{2}(Z) \ \ \ ROC :a_{2}< \left | Z \right | <b_{2}-R_{2}$

$Z\left \{ x_{1}[n]\ x_{2}[n] \right \}\leftrightarrow \ \ ?$ .

we know that $X(Z) = \sum_{n=-\infty }^{\infty } x[n] \ Z^{-n}$ .

$Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(x_{1}[n]\ x_{2}[n]) \ Z^{-n}$ .

$Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(\ x_{2}[n]\ Z^{-n}) \ \frac{1}{2\pi j} \oint_{c}X_{1}(v)\ v^{n-1}\ dv$ .

$Z\left \{ x_{3}[n]\right \} = \sum_{n=-\infty }^{\infty }(\ x_{2}[n]\ (\frac{Z}{v} )^{-n} \ \frac{1}{2\pi j} \oint_{c}X_{1}(v)\ v^{-1}\ dv$ .

$Z\left \{ x_{3}[n]\right \} = \ \frac{1}{2\pi j} \oint_{c}X_{2}(\frac{Z}{v} ) X_{1}(v)\ v^{-1}\ dv$ .

Parseval’s Relation :-

The two complex valued signals $x_{1}[n] , \ x_{2}[n]$ then the Parseval’s relation states that

$\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n] = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{1}{v^{*}} ) v^{-1}\ dv$ .

Proof:-

By using complex convolution theorem

$\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n]\ Z ^{-n} = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{Z^{*}}{v^{*}} ) v^{-1}\ dv$ .

in the above equation substitute $Z=1$ , then

$\sum_{n=-\infty }^{\infty } x_{1}[n] \ x_{2}^{*}[n] = \ \frac{1}{2\pi j} \oint_{c} X_{1}(v)\ X_{2}^{*}(\frac{1}{v^{*}} ) v^{-1}\ dv$ .

Hence Parseval’s relation is proved.

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Initial value & Final value Theorems in Z-Transforms:-

Initial value Theorem (Z-Transforms):-

If uni-lateral Z-Transform of x[n] is $X_{+}(Z)$ then $x(0) = \lim_{z\rightarrow \infty } X_{+}(Z)$ .

$X_{+}(Z) = \sum_{n=0}^{\infty } x[n] \ Z^{-n}$ .

$X_{+}(Z) = x(0) +x(1)\ Z^{-1}+ x(2)\ Z^{-2}+x(3)\ Z^{-3}+......$

$\lim_{Z\rightarrow \infty }X_{+}(Z) = \lim_{Z\rightarrow \infty }\left$

$\lim_{Z\rightarrow \infty }X_{+}(Z) = x(0)$ .

Final value Theorem (Z-Transforms):-

If uni-lateral Z-Transform of x[n] is $X_{+}(Z)$ then $x(\infty ) = \lim_{z\rightarrow 1}(Z-1) \ X_{+}(Z)$ .

$Z\left \{ x(n+1) \right \}-Z\left \{ x(n) \right \} = \lim_{k\rightarrow \infty }\sum_{n=0}^{k} (x[n+1]-x[n]) \ Z^{-n}$ .

$Z\ X_{+}(Z) -x(0)-X_{+}(Z)= \lim_{k\rightarrow \infty }\sum_{n=0}^{k} (x[n+1]-x[n]) \ Z^{-n}$ .

$\lim_{z\rightarrow 1}\left =\lim_{z\rightarrow 1} \lim_{k\rightarrow \infty }\sum_{n=0}^{k} \left$

$\lim_{z\rightarrow 1}\left = \lim_{z\rightarrow 1}\left \{ x(\infty )-x(0) \right \}$ .

$\lim_{z\rightarrow 1}(Z-1)\ X_{+}(Z) = x(\infty )$ .

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Methods to find out inverse Z-transforms

Long division (or) Power Series Expansion Method:-

$X(Z) = \sum_{n=-\infty }^{\infty }x[n]\ Z^{-n}$

$X(Z)$ can be expressed either in positive powers of Z or negative powers of Z.

if the sequence is causal $X(Z)$ has negative powers of Z similarly the Non-causal sequence $X(Z)$ negative powers of Z.

Let $X(Z)=\frac{N(Z)}{D(Z)} =\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}}$ .

$X(Z)$ is causal it has ROC $\left | Z \right |> \left | r \right |$ then $X(Z)$ can be expressed as

$X(Z) = x(0)+x(1)Z^{-1}+x(2)Z^{-2}+x(3)Z^{-3}+........$

$X(Z)$ is non-causal it has ROC $\left | Z \right |<\left | r \right |$ then $X(Z)$ can be expressed as

$X(Z)=x(0)+x(-1)Z^{1}+x(-2)Z^{2}+x(-3)Z^{3}+...........$

Partial fraction Method:-

Let $X(Z)=\frac{N(Z)}{D(Z)} =\frac{b_{0}+b_{1}Z^{-1}+b_{2}Z^{-2}+...........+b_{M}Z^{-M}}{a_{0}+a_{1}Z^{-1}+a_{2}Z^{-2}+...........+a_{N}Z^{-N}}$ and $a_{o}=1$

if $M<N$ , $X(Z)$ is a proper function .

if $M\geq N$ , $X(Z)$ is improper function so convert the improper function to proper function as

$X(Z)=c_{o}+c_{1}Z^{-1}+c_{2}Z^{-2}+.........+c_{M-N}Z^{-(M-N)}+ \frac{N_{1}(Z)}{D(Z)}$ .

$X(Z) = polynomial + rational \ proper \ function$ .

express $X(Z )$ into powers of Z as follows

$X(Z ) = \frac{b_{o}Z^{N}+b_{1}Z^{N-1}+b_{2}Z^{N-2}+.......+b_{M}Z^{N-M}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}$

then divide $X(Z)$ by Z

$\frac{X(Z)}{Z} = \frac{b_{o}Z^{N-1}+b_{1}Z^{N-2}+b_{2}Z^{N-3}+.......+b_{M}Z^{N-M-1}}{Z^{N}+a_{1}Z^{N-1}+a_{2}Z^{N-2}+.......+a_{N}}$ .

Now express $\frac{X(Z)}{Z}$ into partial fractions using different cases and find out the inverse Z-transform for the function

$X(Z ) = Z.(partial fraction \ expansion )$ .

Convolution Method:-

express $X(Z)$ as a product of two functions $X_{1}(Z)$ and $X_{2}(Z)$ as follows $X(Z) =X_{1}(Z) . X_{2}(Z)$

then find the inverse Z- transforms of individual functions

$x_{1}[n]\leftrightarrow X_{1}(Z)$

$x_{2}[n]\leftrightarrow X_{2}(Z)$

by using convolution method find convolution of $x_{1}[n]$ and $x_{2}[n]$

i.e, $x[n] = x_{1}[n] * x_{2}[n]$

now $x[n]$ is the inverse Z-transform of $X(Z)$ .

Cauchy Residue Theorem:-

$f(Z)$ a function in Z if the derivative $\frac{df(Z)}{dZ}$ exists on and inside contour C and $f(Z)$ has no poles at $Z=Z_{o}$ then.

$\frac{1}{2\pi j}\oint_{c} \frac{f(Z)dZ}{Z-Z_{o}}=\left\{\begin{matrix} f(Z_{o}) \ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \end{matrix}\right.$ .

if $(k+1)^{th}$ the derivative of $f(Z)$ exists on and has no poles at $Z=Z_{o}$ then.

$\frac{1}{2\pi j}\oint_{c} \frac{f(Z)}{(Z-Z_{o})^k}dZ=\left\{\begin{matrix}\frac{1}{(k-1)!} \frac{d^{k-1}f(Z)} {dZ^{k-1}}\ if \ Z_{o} \ is\ inside \ C\\ 0 \ if \ Z_{o}\ is\ outside \ C \end{matrix}\right.$ .

the values on the right hand side are called Residue’s of the pole $Z=Z_{o}$ .

if there are n no of poles inside C $\frac{1}{2\pi j}\oint_{c} \frac{f(Z)dZ}{(Z-Z_{1})(Z-Z_{2})(Z-Z_{3})..(Z-Z_{n})}=\sum_{i=1}^{n}\lim_{Z\rightarrow Z_{i}} \left$ .

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Relation between Laplace and Fourier Transform

The Fourier transform of a signal x(t) is given as

$X(j\omega ) = \int_{-\infty }^{\infty } x(t) e^{-j\omega t}dt----EQN(I)$

Fourier Transform exists only if $\int_{-\infty }^{\infty } \left | x(t) \right |dt< \infty$

we know that $s=\sigma + j\omega$

$X(S) = \int_{-\infty }^{\infty } x(t) e^{-s t}dt$

$X(S) = \int_{-\infty }^{\infty } \left | x(t)e^{-\sigma t} \right | e^{-j\omega t}dt----EQN(II)$

if we compare Equations (I) and (II) both are equal when $\sigma =0$ .

i.e, $X(S) =X(j\omega)| \right |_{s=j\omega }$ .

This means that Laplace Transform is same as Fourier transform when $s=j\omega$ .

Fourier Transform is nothing but the special case of Laplace transform where $s=j\omega$ indicates the imaginary axis in complex-s-plane.

Thus Laplace transform is basically Fourier Transform on imaginary axis in the s-plane.

Relation between Laplace and Fourier Transforms

index index5a58 index9a7b index9f21 index62e6 index3081

aliasing effect in Sampling

Effect of under sampling (aliasing effect):-

When a Continuous Time band-limited signal is sampled at, then the successive cycles of the spectrum of the sampled signal overlap with each other as shown below

Some aliasing is produced in the signal this is due to under-sampling.

aliasing is the phenomenon in which a high-frequency component in the frequency spectrum of the signal takes as a low-frequency component in the spectrum of the sampled signal.

Because of aliasing, it is not possible to reconstruct x(t) from g(t) by low pass filtering.

The spectral components are in the overlapping regions and hence the signal is distorted.

Since any information signal contains a large no. of frequencies so the decision of sampling frequency always becomes a problem.

A signal is first passed through LPF before sampling.

i.e, it is band limited by this LPF which is known as a pre-alias filter.

To avoid aliasing

Pre-alias filter must be used to limit the bandwidth of the signal to $f_{m}$ Hz.
Sampling frequency must be $f_{s}>2f_{m}$ .

Pre-alias filter means before sampling is passed through an LPF to make a perfect band-limited signal.

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