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

i. Multiplying S(t) with \cos \omega _{c}t.

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

S_{I}(t) = S(t) \cos \omega _{c}t

By finding the Fourier Transform of in-phase component

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

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

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

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

S_{Q}(t) = S(t) \sin \omega _{c}t

By finding the Fourier Transform of Q-phase component

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

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

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

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


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)


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

since Signum function is 

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

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

s_{Q}(f) = (-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) = \frac{A_{c}}{2}m(t)\cos \omega _{c}t-\frac{A_{c}}{2}\widehat{m(t)}\sin \omega _{c}t.

similarly, SSB signal when only LSB has been transmitted 

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


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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.