# PSD of modulated signal

I know this question has been previously asked just saw it and tried to do myself but reaching an expression which is different from expression of PSD which I remember for modulated signal

Actual answer $$S(f-f_c)+S(f+f_c)$$

My answer $$S(f-f_c)+S(f+f_c)+2X(f-f_c)X(f+f_c)$$

where $$S(f)$$ is PSD of $$x(t)$$ and $$X(f)$$ is its Fourier transform while it is modulated by $$\cos$$ signal of $$f_c$$ frequency.

What I have followed is to find fourier transform of modulated signal and then square it.Where i have done wrong. • If you've got a different result than what is correct, you'll obviously need to give us your full derivation. How else are we supposed to find your mistake? My guess is you just forgot to take a critical step, but I can't know, and since, as you say, there's enough existing derivations out there that you can compare your own steps to, I'm not willing to spend time speculating. – Marcus Müller Jul 8 '19 at 16:08
• If $X(f)$ is a low-pass function (say to 10 kHz(and $f_c$ a high-frequency carrier signal (say 1MHz), can you tell me a specific value of $f$ for which $X(f-f_c)X(f+f_c)$ is nonzero? – Dilip Sarwate Jul 9 '19 at 1:47
• @Dilip Sarwate I know that this term will be zero for low pass input (fc>W) but I no where mentioned x(t) is low pass. – Buzz bee Jul 9 '19 at 1:54
• Congratulations. And while we are nitpicking about the details of what you wrote, let me remind you that the PSD is not the square of the Fourier transform of the signal. – Dilip Sarwate Jul 9 '19 at 11:43
• @DilipSarwate Iconfused ESD with PSD sorry,Then how to I derive PSD of modulated signal its auto-correlation is coming R(a)cos(vt)cos(v(t+a)) where R(a) is auto-correlation of x(t) and v is modulating carrier frequency. – Buzz bee Jul 9 '19 at 14:34

If $$x(t)$$ is a finite-energy signal with Fourier transform $$X(f)$$, then $$x(t)\cos(2\pi f_c t)$$ is also a finite-energy signal with Fourier transform $$\left.\left.\frac 12 \right[X(f-f_c) + X(f+f_c)\right]$$. This is just the modulation theorem of Fourier transform theory.
The energy spectral density of $$x(t)$$ is $$S_x(f) = |X(f)|^2$$ while the energy spectral density of $$y(t) = x(t)\cos(2\pi f_c t)$$ is \begin{align}S_y(f) &= \bigg|\left.\left.\frac 12 \right[X(f-f_c) + X(f+f_c)\right]\bigg|^2\\ &= \left.\left.\frac 14 \right[ |X(f-f_c)|^2 + |X(f+f_c)|^2 + 2\Re\big(X(f-f_c)X^*(f+f_c)\big)\right]\\ &= \left.\left.\frac 14 \right[S_x(f-f_c)+S_x(f+f_c)+2\Re\big(X(f-f_c)X^*(f+f_c)\big)\right]\tag{1} \end{align} which has a vague resemblance to what the OP wrote. Now, if $$x(t)$$ is a low-pass signal with bandwidth $$W$$ Hz and $$W < f_c$$, then $$X(f-f_c)$$ occupies the frequency band from $$f_c-W > 0$$ to $$f_c+W$$ while $$X(f+f_c)$$ occupies the frequency band from $$-f_c-W$$ to $$-f_c+W < 0$$. We see that there is no value of $$f$$ for which $$X(f-f_c)$$ and $$X(f+f_c)$$ both are nonzero, and thus $$X(f-f_c)X^*(f+f_c)$$ equals $$0$$ for all $$f$$. We are left with \begin{align} S_y(f) &= \left.\left.\frac 14 \right[S_x(f-f_c)+S_x(f+f_c)\right]\\ &= \left.\left.\frac 14\right[|X(f-f_c)|^2+|X(f+f_c)|^2\right].\tag{2}\end{align} But when $$f_c$$ is smaller than $$W$$, then, as the OP peevishly has informed us all that he has placed no restrictions on the value of $$f_c$$, he must learn to live with the more complicated $$(1)$$.
• In modulation there is no local oscillator and a separate carrier frequency to create a sum and difference frequency. Alternatively, the local oscillator is at zero frequency and so the sum and difference frequencies are $f_c$ and $-f_c$, i.e. the spectrum moves to be centered at $\pm f_c$, where $f_c$ is the carrier frequency. – Dilip Sarwate Jul 9 '19 at 11:48