# Finding causal impulse response given the imaginary part of the frequency response I understand that I would need to calculate inverse Discrete Time Fourier Transform (iDTFT) to find $$h(n)$$. Since $$h(n)$$ is real, iDTFT of the imaginary part of $$H(e^{j\omega})$$ gives the odd part of h(n) and iDTFT of the real part of $$H(e^{j\omega})$$ part would give the even part of $$h(n)$$ (Correct me if I am wrong here). But, how do I find the even part of $$h(n)$$ when I only have imaginary part of frequency response and find all the possible $$h(n)$$?

You're right about the relation between the imaginary part of the frequency response and the odd part of the impulse response:

$$\textrm{DTFT}\big\{h_o[n]\big\}=jH_I(e^{j\omega})\tag{1}$$

So from the given $$H_I(e^{j\omega})$$ you can obtain $$h_o[n]$$. Now note that the odd part of $$h[n]$$ is defined by

$$h_o[n]=\frac12\big(h[n]-h[-n]\big)\tag{2}$$

The important extra information that you have is that the system is causal. This means that $$h[n]=0$$ for $$n<0$$. Consequently, from $$(2)$$ we obtain

$$h[n]=\begin{cases}2h_o[n],&n>0\\0,& n<0\end{cases}\tag{3}$$

The only ambiguity we have is the value of $$h$$, which cannot be obtained from the imaginary part of the frequency response.

Note that the same solution can be obtained by using the fact that for causal signals, the real and imaginary parts of the frequency response are related via the Hilbert transform. But the above method is probably even more intuitive and doesn't require any knowledge of the Hilbert transform.