# Time-domain representation of SSB-SC signal

Let the signal produced by SSB-SC modulator is S(t), a Band pass signal

$S_{USB}(t)=S_{I}(t)&space;cos&space;\omega&space;_{c}t-S_{Q}(t)&space;sin&space;\omega&space;_{c}t$, where $S_{I}(t)$ is  In-Phase component of S(t) obtained by

i. Multiplying S(t) with $\cos&space;\omega&space;_{c}t$.

ii. and passing the product through a LPF with suitable cut-off frequency.

$S_{I}(t)&space;=&space;S(t)&space;\cos&space;\omega&space;_{c}t$

By finding the Fourier Transform of in-phase component

$S_{I}(f)&space;=&space;\frac{1}{2}(S(f-f_{c})+S(f+f_{c}))$

after restricting the signal $S_{I}(f)$ between   $-B\leq&space;f&space;\leq&space;B$

$S_{I}(f)&space;=\left\{\begin{matrix}&space;\frac{1}{2}(S(f-f_{c})+S(f+f_{c})),-B\leq&space;f&space;\leq&space;B&space;&&space;\\0&space;,&space;otherwise&space;&&space;\end{matrix}\right.$

Similarly $S_{Q}(t)$ is the quadrature phase component of s(t), obtained by multiplying S(t) with $\sin&space;\omega&space;_{c}t$  and by passing the resultant signal through a LPF .

$S_{Q}(t)&space;=&space;S(t)&space;\sin&space;\omega&space;_{c}t$

By finding the Fourier Transform of Q-phase component

$S_{Q}(f)&space;=&space;\frac{1}{2j}(S(f-f_{c})-S(f+f_{c}))$

after restricting the signal $S_{Q}(f)$ between   $-B\leq&space;f&space;\leq&space;B$

$S_{Q}(f)&space;=\left\{\begin{matrix}&space;\frac{1}{2j}(S(f-f_{c})-S(f+f_{c})),-B\leq&space;f&space;\leq&space;B&space;&&space;\\0&space;,&space;otherwise&space;&&space;\end{matrix}\right.$

Now Let’s assume S(f) is the required frequency spectrum of SSB-SC signal when only USB has been transmitted.

i.e,

from the above figure,

one can obtain $S(f-f_{c})$ ,  $S(f+f_{c})$ by shifting the signal S(f) towards right by $f_{c}$  and  left by  $f_{c}$

Now by adding  $S(f-f_{c})$  and  $S(f+f_{c})$

from the above figure, $S_{I}(f)$ results to be

from the frequency spectrum of $S_{I}(f)$ , the time-domain representation turns out to be $S_{I}(t)=\frac{1}{2}A_{c}m(t)-----EQN(I)$

Similarly,

The resultant signals $S(f-f_{c})-S(f+f_{c})$ and $S_{Q}(f)$

from the frequency spectrum of $S_{Q}(f)$ turns out to be

$S_{Q}(f)&space;=\left\{\begin{matrix}&space;\frac{1}{2j}A_{c}M(f),-B\leq&space;f&space;\leq&space;0&space;&&space;\\&space;-\frac{1}{2j}A_{c}M(f),0\leq&space;f&space;\leq&space;B&space;&&space;\end{matrix}\right.$

since Signum function is

$Sign(f)&space;=\left\{\begin{matrix}&space;+1,f>0&space;&&space;\\&space;-1&space;f<0&space;&&space;\end{matrix}\right.$

$S_{Q}(f)$ when expressed in terms of Signum function $s_{Q}(f)&space;=&space;\frac{j}{2}A_{c}M(f)(-sign(f))$

$s_{Q}(f)&space;=&space;(-jsign(f)M(f))\frac{A_{c}}{2}$

By using Hilbert transform of m(t) , the time-domain representation turns out to be $S_{Q}(t)=\frac{1}{2}A_{c}\widehat{m(t)}-----EQN(II)$

From EQN’s (I) and (II) , the time-domain representation of SSB-SC signal results

$S_{USB}(t)&space;=&space;\frac{A_{c}}{2}m(t)\cos&space;\omega&space;_{c}t-\frac{A_{c}}{2}\widehat{m(t)}\sin&space;\omega&space;_{c}t$.

similarly, SSB signal when only LSB has been transmitted

$S_{LSB}(t)&space;=&space;\frac{A_{c}}{2}m(t)\cos&space;\omega&space;_{c}t+\frac{A_{c}}{2}\widehat{m(t)}\sin&space;\omega&space;_{c}t$

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