Title | First Order Frequency Response |
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Author | Abishake Vimalan |
Course | Systems Modelling, Simulation and Computation |
Institution | The University of Warwick |
Pages | 4 |
File Size | 269.2 KB |
File Type | |
Total Downloads | 67 |
Total Views | 138 |
Additional notes, worked examples and practice materials on Gain and Phase for systems modelling...
Gain and Phase Gain is the ratio of output to input amplitude, ie how much bigger or smaller the output is compared to input.
Phase difference ϕ(ω) is how much the output signal is delayed compared to the input signal. Both are functions of input frequency ω. The frequency response can be obtained by substituting jω for s in the transfer function. This gives a complex function as shown
Magnitude ∣G(jω)∣ gives the amplitude of the response, and the argument of the complex number ∠G(jω) gives the phase shift ϕ. The substitution s=jω is used, is because in the Laplace domain, both signals and systems are represented by functions of s.
The s-plane is the complex plane on which Laplace transforms are graphed. Generally, s=σ+jω σ is the Neper frequency, the rate at which the function decays ω is the radial frequency, the rate at which the function oscillates
Periodic sinusoidal inputs are non decaying, so σ=0, giving s=jω
To find the frequency response parameters:
The graphs below show the frequency response in terms of T for varying frequency ω:
Example Given a transfer function G=1/s, what is the magnitude and phase of frequency response
Bode Plots Bode plots show frequency and amplitude of frequency response on a log10 scale. Information is not spread linearly accross the frequency range, so it makes more sense to use a logarithmic scale. An important feature of bode plots is the corner frequency: the frequency at the point where the two asymptotes of the magnitudefrequency graph. This point is where ω=1/T
The plot above is for the function G(s)=1/(s+1). The gain is measured in decibels dB for the magnitude of the response....