3 deleted 72 characters in body edited May 17 '13 at 10:24 Matt L. 55.5k22 gold badges4040 silver badges103103 bronze badges You're not telling us where this pops up but my some magic I believe to know thatsuppose you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $$h(t)$$: $$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$ Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $$X(j\omega)$$ is the Fourier transform of the input signal, and $$Y(j\omega)$$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$ For a sinusoidal input signal $$x(t)=\sin (\omega_0 t)$$ the output signal $$y(t)$$ is given by $$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$ where $$\text{arg}\{H(j\omega_0\})$$ is the phase of the complex function $$H(j\omega)$$ at frequency $$\omega_0$$. So you can see that $$|H(j\omega)|$$ represents the amplification of the system at frequency $$\omega$$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $$H(j\omega)$$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works. You're not telling us where this pops up but my some magic I believe to know that you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $$h(t)$$: $$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$ Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $$X(j\omega)$$ is the Fourier transform of the input signal, and $$Y(j\omega)$$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$ For a sinusoidal input signal $$x(t)=\sin (\omega_0 t)$$ the output signal $$y(t)$$ is given by $$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$ where $$\text{arg}\{H(j\omega_0\})$$ is the phase of the complex function $$H(j\omega)$$ at frequency $$\omega_0$$. So you can see that $$|H(j\omega)|$$ represents the amplification of the system at frequency $$\omega$$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $$H(j\omega)$$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works. I suppose you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $$h(t)$$: $$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$ Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $$X(j\omega)$$ is the Fourier transform of the input signal, and $$Y(j\omega)$$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$ For a sinusoidal input signal $$x(t)=\sin (\omega_0 t)$$ the output signal $$y(t)$$ is given by $$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$ where $$\text{arg}\{H(j\omega_0\})$$ is the phase of the complex function $$H(j\omega)$$ at frequency $$\omega_0$$. So you can see that $$|H(j\omega)|$$ represents the amplification of the system at frequency $$\omega$$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $$H(j\omega)$$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works. 2 deleted 1 characters in body edited May 17 '13 at 10:14 Matt L. 55.5k22 gold badges4040 silver badges103103 bronze badges You're not telling us where this pops up but my some magic I believe to know that you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $$h(t)$$: $$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$ Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $$X(j\omega)$$ is the Fourier transform of the input signal, and $$Y(j\omega)$$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$ For a sinusoidal input signal $$x(t)=\sin (\omega_0 t)$$, the output signal $$y(t)$$ is given by $$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$ where $$\text{arg}\{H(j\omega_0\})$$ is the phase of the complex function $$H(j\omega)$$ at frequency $$\omega_0$$. So you can see that $$|H(j\omega)|$$ represents the amplification of the system at frequency $$\omega$$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $$H(j\omega)$$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works. You're not telling us where this pops up but my some magic I believe to know that you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $$h(t)$$: $$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$ Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $$X(j\omega)$$ is the Fourier transform of the input signal, and $$Y(j\omega)$$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$ For a sinusoidal input signal $$x(t)=\sin (\omega_0 t)$$, the output signal $$y(t)$$ is given by $$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$ where $$\text{arg}\{H(j\omega_0\})$$ is the phase of the complex function $$H(j\omega)$$ at frequency $$\omega_0$$. So you can see that $$|H(j\omega)|$$ represents the amplification of the system at frequency $$\omega$$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $$H(j\omega)$$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works. You're not telling us where this pops up but my some magic I believe to know that you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $$h(t)$$: $$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$ Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $$X(j\omega)$$ is the Fourier transform of the input signal, and $$Y(j\omega)$$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$ For a sinusoidal input signal $$x(t)=\sin (\omega_0 t)$$ the output signal $$y(t)$$ is given by $$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$ where $$\text{arg}\{H(j\omega_0\})$$ is the phase of the complex function $$H(j\omega)$$ at frequency $$\omega_0$$. So you can see that $$|H(j\omega)|$$ represents the amplification of the system at frequency $$\omega$$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $$H(j\omega)$$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works. 1 answered May 17 '13 at 10:09 Matt L. 55.5k22 gold badges4040 silver badges103103 bronze badges You're not telling us where this pops up but my some magic I believe to know that you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $$h(t)$$: $$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$ Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $$X(j\omega)$$ is the Fourier transform of the input signal, and $$Y(j\omega)$$ is the Fourier transform of the output signal, then $$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$ For a sinusoidal input signal $$x(t)=\sin (\omega_0 t)$$, the output signal $$y(t)$$ is given by $$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$ where $$\text{arg}\{H(j\omega_0\})$$ is the phase of the complex function $$H(j\omega)$$ at frequency $$\omega_0$$. So you can see that $$|H(j\omega)|$$ represents the amplification of the system at frequency $$\omega$$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $$H(j\omega)$$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works.