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For an LTI system with bounded input and bounded output, I have the input $$x(t) = 5 + \cos(12t+\pi/4)$$ and output $$y(t) = 6\sin(12t)$$

It is said that the magnitude of the frequency response $|H(j12)| = 6$. I don't understand how this is true. Obviously the second term of $x(t)$ is scaled by $6$, but the first term needs to be scaled by $0$, right? Which means this LTI system isn't possible. Can anyone clarify?

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For a sinusoidal input $x(t)=\sin(\omega_0t+\varphi)$, the output of a real-valued LTI system is given by

$$y(t)=|H(j\omega_0)|\sin(\omega_0t+\varphi+\arg\{H(j\omega_0)\})\tag{1}$$

where $H(j\omega)$ is the system's frequency response. The amplitude of the output is scaled by the magnitude of $H(j\omega)$ evaluated at $\omega=\omega_0$, and the phase is shifted by the argument of $H(j\omega_0)$.

If the input is a constant ($x(t)=c$), then in analogy with $(1)$, the output is given by

$$y(t)=|H(j0)|\cdot c\tag{2}$$

Since the system is linear, we can use superposition, and compute the output for each input component and then sum up all the individual components. So the output corresponding to the input $x(t)=c+\sin(\omega_0t+\varphi)$ is given by

$$y(t)=|H(j0)|\cdot c+|H(j\omega_0)|\sin(\omega_0t+\varphi+\arg\{H(j\omega_0)\})\tag{3}$$

So there is no contradiction between a gain of $6$ at frequency $\omega_0$ and a gain of $0$ at frequency $0$ (DC). From the given input and output you can deduce the values of $H(0)$ and $H(\omega_0)$.

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  • $\begingroup$ Thanks I forgot about the fact that H(jw) was evaluated differently $\endgroup$ – Goldname Feb 5 '17 at 16:46
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It is a stable system. Stability depends on the amplitude of the output. Since for a bounded input (Values varying between +6 and +4) the output is varying between -6 and +6 (Bounded output).

The value 5 may or may not be scaled. See the below example. It might give an insight.

Say we are passing the value 5 into a system as below.

System H(jw) = some w*y(w). Convolution in time domain is multiplication in frequency domain. Fourier Transform of 5 is 10*π*𝛿(w). Which when passed through the system gives output as zero as w*𝛿(w) = 0. Here also the system is said to be stable as input and output both are bounded again.

Also note that even a million examples might not a prove a system as stable. But a single counter example can prove a system as unstable.

Here in your question as system function is not given and I/P O/P relation is only given, we have to go by BIBO principle applying on I/P and O/P.

Hope this helps you....

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Hint. The output of a Linear system in front of a sum of inputs is (by libearity) the sum of the output in front of the single inputs. Thus the frequency response of the system just have to be zero at $\omega=0$ and scaled by 6 (and delayed by $\pi/4$) at $\omega=12$. Any idea?

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