# How do I take the real part of this bandpass filter system's output?

I'm stuck on a final step in this problem. Essentially, there are the two systems above, which we'll call System 1 (Fig. 4.26, with ideal lowpass $H(jw)$) and System 2 (with $H_1(jw)$). The question proposes that the output y(t) of System 2 is identical to that which would be obtained by retaining $\Re\{f(t)\}$ of System 1.

I have a couple questions -

(1) Why do we need to take the real part of system 1 and not system 2? Aren't they both already bandpass filters?

(2) I solved out answers for $F(jw)$ and $Y(jw)$. $H_1(jw)=H(jw)$ is assumed (they seem to be the same):

$$F(jw)=H(j(w+w_c))X(jw)\\Y(jw)=\frac{1}{2}(H(j(w-w_c))X(jw)+\frac{1}{2}H(j(w+w_c))X(jw)$$

I'm relatively confident this answer is correct, but I'm unsure of how to take the real part of $f(t)$. Can I take the real part of $F(jw)$ and get it to look like $Y(jw)$?

Thank you!

• Hello can you know this question from any book – king love Oct 27 '18 at 6:07

The signal $f(t)$ is a complex bandpass signal. Note that if $H(e^{j\omega})$ is an ideal lowpass filter (and $\omega_c>\omega_0$) then $f(t)$ only has positive frequency components. For a real-valued signal, the spectrum shows conjugate symmetry, so there have to be negative frequency components as well.
By taking the real part of $f(t)$ you get $y(t)$:
$$\Re\{f(t)\}=\frac12\left[f(t)+f^*(t)\right]\Longleftrightarrow \frac12\left[F(j\omega)+F^*(-j\omega)\right]=\\=\frac12\left[X(j\omega)H(j(\omega+\omega_c))+X^*(-j\omega)H^*(j(\omega_c-\omega))\right]\tag{1}$$
(where '$\Longleftrightarrow$' means Fourier transform pair).
Since $x(t)$ and $h(t)$ (the impulse response of the ideal lowpass filter) are real-valued, their Fourier transforms satisfy $X(j\omega)=X^*(-j\omega)$ and $H(j\omega)=H^*(-j\omega)$. Using this property, Eq. (1) can be rewritten as
$$\Re\{f(t)\}\Longleftrightarrow\frac12 X(j\omega)\left[H(j(\omega+\omega_c))+H(j(\omega-\omega_c))\right]=Y(j\omega)$$