I am having difficulty with this question specifically with the $\sin()$ part. So here goes nothing:

Consider a Linear Time-Invariant system with input $x[n]$ and output $y[n]$. When the input to the system is $$x[n]=5 \sin(0.4\pi n)/(\pi n) + 10 \cos(0.5 \pi n)$$ and corresponding output is $$y[n] = 10 \sin(0.3 \pi (n-10))/(\pi (n-10))$$

Determine the frequency response $H(e^{jw})$ and the impulse response $h[n]$ for the Linear Time-Invariant system.

Thanks in advance.

  • 1
    $\begingroup$ It would help to know what LTI means $\endgroup$ Commented Feb 28, 2016 at 21:50
  • $\begingroup$ You'll have a better chance of getting help if you show what you have tried, and exactly where you're having trouble making progress. $\endgroup$
    – MBaz
    Commented Feb 29, 2016 at 3:37

1 Answer 1


A classical question on signals and systems, for which I provide you one method of solution.

Now given an LTI system, and its input $x[n]$ and output $y[n]$, the fundamental approach to determine the impulse response $h[n]$ involves the utilisation of a frequency domain approach to assert $H(e^{jw}) = Y(e^{jw})/X(e^{jw})$

In this problem however, I would choose a different approach which involves more of an analysis by inspection rather than pure mathematical operations, because of the fact that the input $x[n]$ involves a cosine term whose DTFT is includes impulses and will be problematic in the division process.

The method proceeds by drawing the DTFTs of input and output: The input $x[n]$ involves two terms one of which is $$5 \sin(0.4 \pi n)/ \pi n$$ whose DTFT is a zero phase ideal lowpass filter of cutoff frequency $W_c = 0.4 \pi$ with a gain of 5.

And the second term of input is $$10 \cos (0.5 \pi n)$$ whose DTFT is modeled as two impulses at $$ 5 \pi \delta(\omega + 0.5 \pi) + 5 \pi \delta(\omega - 0.5 \pi) $$ Because of linearity property of DTFT, these two DTFT are added to represent the DTFT of the input $x[n]$.

Then, look at the DTFT of the output $$y[n] = \frac {10 \sin(0.3 \pi (n-10) )}{\pi (n-10)} $$ which is again an ideal low pass filter of gain = 10 and $W_c = 0.3 \pi$ with a phase term of $e^{-10 \omega}$ hence $Y(e^{j\omega}) =10 e^{-j10 \omega}$ when $|\omega| < 0.3 \pi$ and zero elsewhere.

Now we want to find a frequency response $H(e^{j\omega})$ such that when it is multiplied by the DTFT $X(e^{j\omega})$ of the input $x[n]$, it should produce the DTFT $Y(e^{j\omega})$ of the output $y[n]$.

The observation of the input and output DTFTs yields the conclusion that one such $H(e^{j \omega})$ is a linear phase ideal lowpass filter of gain = 2, $W_c = 0.3\pi$ and phase term of $e^{-j10 \omega}$, whose time domain correspondence is found by inspection as $$h[n] = \frac {2 \sin(0.3 \pi (n-10))}{\pi (n-10)}$$

The justification of this solution can be made by observing the effect of this filter on its input: it will completely supress any signal above the frequency $\omega = 0.3 \pi$ and it will multiply any signal below that frequency by 2, while adding a phase shift of -10 radians which reflects itself as a group delay of 10 samples in (n-10) term.


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.