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Working on this problem on linear invariant systems in signal processing, but unsure if I've got the right answer:

Show that for a real impulse response function of $H(\omega)$, the response to a sine input of $\sin(\omega_0 t)$ is the imaginary part of $H(\omega_0)e^{i\omega_0 t}$.

My answer:

Let's assume the input $f(t)$ is equal to $\sin(\omega_0 t)$. Expressing $\sin(\omega_0 t)$ in an exponential form, we have:

$$\sin(\omega_0 t)=\dfrac{e^{i\omega_0 t}-e^{-i\omega_0 t}}{2i}.$$

So the output will be as follows: $$\dfrac1{2i}(H(\omega_0)e^{i\omega_0 t}-H(-\omega_0)e^{-i\omega_0 t}),$$ where we have used the fact that the output of $e^{i\omega_0 t}$ is $H(\omega_0)e^{i\omega_0 t}$.

Using the identity $H(\omega_0)=H^*(-\omega_0)$ (since real IRF), we end up with output as: $$\dfrac{1}{i}(\mbox{Im}(H(\omega_0)e^{i\omega_0 t})$$

I think I'm being a bit silly, but is that it?

Sina
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George93
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  • If $z = x+iy$, do you also express this as $$z = \operatorname{Re}(z) + i\operatorname{Im}(z)?\tag{1}$$ or as $$z = \operatorname{Re}(z) + \operatorname{Im}(z)?\tag{2}$$ The question you are asked assumes the former convention whereas your answer (where you are dividing by $i$) is using the latter convention (Actually I am not sure that the latter convention is at all in common use). – Dilip Sarwate Jan 05 '15 at 14:42

1 Answers1

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I checked your steps, they are correct except the last one. I think the last step should be:

$$y(t)=\frac{1}{2i}[H(\omega_o)e^{i\omega_o t}-H(-\omega_o)e^{-i\omega_ot}]$$

For $h(t)$ is real then $H(t)$ has conjugate symmetry property: $H(\omega_o)=H^*(-\omega_o)$. So

$$y(t)=\frac{1}{2i}[H(\omega_o)e^{i\omega_o t}-H^*(\omega_o)e^{-i\omega_ot}] =\frac{1}{2i}[2i Im\{H(\omega_o)e^{i\omega_ot}\}] =Im\{H(\omega_o)e^{i\omega_ot}\}$$

Jihad
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Bob Yang
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  • I'm missing the i in [2iIm{H...}], is that just because we need to include it since taking Im{} removes it? – George93 Jan 05 '15 at 16:33
  • The complex number z can be expressed in Cartesian form also called rectangular form: z=x+iy, where x is the 'real part' and y is the 'imaginary part'. Imaginary part often use the notation y=Im{z}. Imaginary part is real number and it does not include i. – Bob Yang Jan 06 '15 at 00:41