Imagine we have two signals in the frequncy domain X(f) and H(f), for one of the signal H(f) I have the complex pair and the frequncy at which that complex pair exists. The second could be any random signal with a single frequency in MATLAB with its time domain signal represented in MATLAB as

x = A*sin(2*pi*fc*t + phi)

now we if we want to multiply both these signals for a known frequency lets suppose 100Hz then multiplication simply becomes.

H(jw) * X(jw) = H(j*(2*pi*100)) * X(j*(2*pi*100))

where w = 2pif , its clear that w is same in both these frequencies Now if signal has multiple frequency components

x = A*sin(2*pi*fc1*t + phi) + A*sin(2*pi*fc2*t + phi) + A*sin(2*pi*fc3*t + phi)

and I have complex pairs of the signal for fc1 , fc2, fc3 , how does the multiplication exactly occur? like how does the multiplication makes sure same frequency components are multiplied in each signal , this is bit confusing , maybe I am missing something. My second question is I can reconstruct the signal of H(f) by taking the frequncy phase shift and the magnitude calculated through the complex pair add them and up in time domain and do a convolution in time domain by doing.

g  = conv(x,h)

Is there any other way ?


1 Answer 1


Multiplication in the frequency domain is equivalent to convolution in the time domain.

When the signal occupies multiple frequencies, describe the signal in the frequency domain and then do the product directly. If the component in one signal is zero at any given frequency, the result for that frequency will simply be zero.

So for the example:

$$x_1(t) = A_1\sin(2\pi f_{c1}t + \phi_1) + A_2\sin(2\pi f_{c2}t + \phi_2) + A_3\sin(2\pi f_{c3}t + \phi_3)$$

If this were convolved in time with another signal:

$$x_2(t) = A_4\sin(2\pi f_{c1}t)$$

Then the product in frequency would result in the time domain waveform:

$$x_3(t) = A_1A_4 \sin(2\pi f_{c1}t +\phi_1)$$

Typically when multiplying a complete spectrum in the frequency domain, we would consider the spectrum as a continuous function of frequency rather than describing it with individual sinusoidal components, such as the product of the two spectrums given in the figure below. The third spectrum being the product of the upper two:

spectrum product


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.