Given the following block diagram, find the frequency responses $H1(f)$ and $H2(f)$. The frequency response of the whole system has to be $H(f)=(\alpha_0+\alpha_1e^{-j2\pi T_1f}+\alpha_2e^{-j2\pi T_2f})^{-1}$

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The fact that there's a loop confuses me. I would express $Y(f)$ as $Y(f)=X(f)H1(f)+(X(f)-X(f)H1(f)H2(f))\ H1(f)+...$, but it doesn't seem to be correct. Could you give me some hints? Thanks in advance!

Solution: $H1(f)=\alpha_0^{-1},\ H2(f)=\alpha_0^{-1}\alpha_1e^{-j2\pi T_1f}+\alpha_0^{-1}\alpha_2e^{-j2\pi T_2f}$.

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    $\begingroup$ Hint: the response of feedback system is the forward gain divided by (1 + the loop gain) where the negative feedback is already implied. H1(f) is the forward rain and H1(f)H2(f) is the loop gain. You can derive that equation easily knowing Y(f)= H1(f)(X(f)-Y(f)H2(f))—- Solve that for Y(f)/X(f) $\endgroup$ Jan 12, 2020 at 16:48

1 Answer 1


With feedback systems such as the one given it's often easy to define an additional signal at the output of the adder. This gives the following equations:

$$U(f)= X(f)-H_2(f)Y(f)\tag{1}$$



Now you can solve Eqs $(1)$ and $(2)$ to get the frequency response $H(f)=Y(f)/X(f)$.

  • $\begingroup$ So I have that $Y=H1(X-YH2)=XH1-YH1H2$, then $Y(1+H1H2)=XH1$ so $\frac{Y}{X}=\frac{H1}{1+H1H2}=H$, and a possible solution could be $H1=1$ and $H2=H^{-1}-1$. The proposed solution in the exercise also works. Thanks for helping!! $\endgroup$
    – Gibbs
    Jan 12, 2020 at 17:05
  • $\begingroup$ @Gibbs: that's right! $\endgroup$
    – Matt L.
    Jan 12, 2020 at 17:45

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