Frequency Response
When analyzing linear time-invariant systems (LTI systems) it is often easier to analyze it on the frequency domain. What does this mean? Suppose we apply a sine wave signal into an LTI system, we would get as output another sine wave with the same frequency but with a different amplitude and a different phase angle. Why is this important? In 1807 Joseph Fourier introduced the Fourier series showing that you could write any periodic function as a sum of sine waves with different frequencies, this idea was later extended to non-periodic signals too by the fourier transform so understanding how a system alters the frequency components of a signal can help us understand how the output of a system will relate to it's input.
Contents
Definitions
Suppose we have a system that for a sinusoidal input of angular frequency \(\omega = 2\pi f\) that has an amplitude of \(M_i(\omega)\) and a phase angle of \(\phi_i(\omega)\), the output of our signal will be a sinusoidal wave with frequency \(\omega\), an amplitude of \(M_o(\omega)\) and a phase angle of \(\phi_o(\omega)\) as seen on the picture.We define the magnitude of the frequency response as:
\[M(\omega) = \frac{M_o(\omega)}{M_i(\omega)}\]
And we define the phase angle of the frequency response as:
\[\phi(\omega) = \phi_o(\omega) - \phi_i(\omega)\]
With that we can define the frequency response to be the phasor:
\[G(j\omega) = M(\omega) \angle{\phi(\omega)}\]
Where \(j = \sqrt{-1}\).
Those definitions are very helpful for us because it helps us interpret the system as a multiplication by a complex number.
Suppose we have a system with a transfer function of \(G(S)\), to calculate the frequency response to this system one may simply evaluate: \[G(j\omega) = \lim_{s \rightarrow j\omega} G(S)\]
Consider a system with transfer function \(G(S)\) and a signal \(x(t) = A cos(\omega t) + B sin(\omega t)\) and let \(Y(S) = X(S)G(S)\). By calculating \(\mathscr{L}\{x(t)\}\) we get that:\[Y(S) = \frac{As + B\omega}{s^2 + \omega^2} \cdot G(S) = \frac{As + B\omega}{(s + j\omega)(s - j\omega)} \cdot G(S)\]
\[Y(S) = \left(\frac{K_1}{s + j\omega} + \frac{K_2}{s - j\omega}\right) \cdot G(S)\]
We can now work out the partial fractions decomposition by the limit method where:
\[K_1 = \lim_{s \rightarrow -j\omega} \frac{As + B\omega}{s - j\omega} = {1 \over 2}(A + jB)G(-j\omega) = {1 \over 2}M_x e^{-j\phi_x} M_g e^{-j\phi_g}\] \[K_1 = \frac{M_x M_g}{2} e^{-j(\phi_x + \phi_g)}\]
\[K_2 = \lim_{s \rightarrow j\omega} \frac{As + B\omega}{s + j\omega} = {1 \over 2}(A - jB)G(j\omega) = {1 \over 2}M_x e^{j\phi_x} M_g e^{j\phi_g}\] \[K_2 = \frac{M_x M_g}{2} e^{j(\phi_x + \phi_g)} = K_1^*\]
Where \(M_g = |G(j\omega)|\) and \(\phi_g = \angle{G(j\omega)}\).
\[Y(S) = \frac{\frac{M_x M_g}{2} e^{-j(\phi_x + \phi_g)}}{s + j\omega} + \frac{\frac{M_x M_g}{2} e^{j(\phi_x + \phi_g)}}{s - j\omega}\]
By applying the inverse Laplace transform we have that:
\[y(t) = M_x M_g \left( \frac{e^{j(\omega t + \phi_x + \phi_g)} + e^{-j(\omega t + \phi_x + \phi_g)}}{2} \right) = M_x M_g cos(\omega t + \phi_x + \phi_g)\]
Which could as well be written as the phasor multiplication:
\[M_y\angle{\phi_y} = (M_g\angle{\phi_g})(M_x\angle{\phi_x})\]
Since \(M_g\angle{\phi_g}\) is the phasor representation of \(G(j\omega)\) it follows that a system with transfer function \(G(s)\) has a frequency response \(G(j\omega)\).
What is the frequency response of the following RC circuit?
We start by applying Kirchhoff's voltage law, which follows that: \[Ri(t) + {1 \over C}\int i(t) dt = V_{in}(t)\]By applying the Laplace transform: \[I(S)(R + {1 \over {Cs}}) = V_{in}(S)\] \[\frac{I(s)}{V_{in}(s)} = \frac{Cs}{RCs + 1}\]
The current on the capacitor can also be described as: \[i(t) = C{{dv_{out}(t)} \over {dt}}\] \[I(s) = Cs V_{out}(s)\]
So, the transfer function of this system is: \[\frac{Cs V_{out}(s)}{V_{in}(s)} = \frac{Cs}{RCs + 1} \Rightarrow \frac{V_{out}(s)}{V_{in}(s)} = \frac{1}{RCs + 1} = \frac{\frac{1}{RC}}{s + \frac{1}{RC}}\]
So the frequency response is given by: \[G(j\omega) = \frac{\frac{1}{RC}}{j\omega + \frac{1}{RC}}\]
We can also write it as the phasor: \[G(j\omega) = |G(j\omega)| \angle{tg^{-1}\left(\frac{\mathfrak{Im}\{G(j\omega)\}}{\mathfrak{Re}\{G(j\omega)\}}\right)}\]
Bode plot
References
- Krishnavedala, O. Butterworth response. Retrieved from https://en.wikipedia.org/wiki/Frequency_response