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- Exponential Input; Gain and Phase Lag | Unit I: First Order Differential Equations | Differential Equations | Mathematics | MIT OpenCourseWare
1 Explain the method of optimism to solve this first order ODE
1.1 Front
Explain the method of optimism to solve this first order ODE
\(\dot{x} + 2x = 4e^{3t}\)
1.2 Back
The inspiration is based on the fact that \(\dv{t} e^{rt} = re^{rt}\). As the right hand side (input) is an exponential, maybe the output signal \(x(t)\) is also an exponential function.
Let’s try: \(x_p(t) = Ae^{3t}\), that is not the general solution, it is just one particular solution. We don’t know \(A\) but, we can substitute this particular solution in the ODE and get \(A\).
- \(\dot{x_p} + 2x_p = 3Ae^{3t} + 2Ae^{3t} = 5Ae^{3t}\)
- \(5Ae^{3t}=4e^{3t} \implies A = 4/5\)
- So, \(x_p(t) = 4/5 e^{3t}\)
Also, we know the homogeneous solution of \(\dot{x} + 2x = 0\), \(x_h(t) = Ce^{-2t}\). So, using superposition principle, we can get the general solution
\({\displaystyle x(t) = x_p(t) + x_h(t) = \frac{4}{5} e^{3t} + Ce^{-2t}}\)
2 How can we get the general solution of a first order constant coefficient DE with sinusoidal input?
2.1 Front
How can we get the general solution of a first order constant coefficient DE with sinusoidal input?
\(\dot{x} + kx = B\cos(\omega t)\)
Assuming \(k\), \(\omega\) and \(B\) are all positives
2.2 Back
The strategy is to use Euler’s formula to replace \(\cos(\omega t)\) by the complex exponential \(e^{i\omega t}\). We call this technique complex replacement
Complex replacement: \(\dot{z} + kz = Be^{i \omega t}\), where \(\cos(\omega t) = \operatorname{Re}(e^{i\omega t})\) and \(x(t) = \operatorname{Re}(z)\)
Using the optimistic method of exponential input will get an exponential solution. Using the particular solution \(z_p(t) = Ae^{i \omega t}\)
- \(\dot{z_p} + k z_p = i \omega A e^{i\omega t} + kAe^{i\omega t} = (k + i\omega)Ae^{i\omega t}\)
- \((k + i\omega)Ae^{i\omega t} = Be^{i\omega t} \implies A = \frac{B}{k + i\omega}\)
So \(z_p(t) = \frac{B}{k + i\omega} e^{i\omega t}\), using polar coordinates \(z_p(t) = \frac{B e^{i(\omega t - \phi)}}{\sqrt{k^2 + \omega^{2}}}\). Where \(\phi = \operatorname{Arg}(k + i\omega)\)
Taking the real part: \({\displaystyle x_p(t) = \frac{B}{\sqrt{k^2 + \omega^2}} \cos(\omega t - \phi)}\).
Finally, the general solution is got using homogeneous solution and principle of superposition
\({\displaystyle x(t) = x_p(t) + Ce^{-kt} = \frac{B}{\sqrt{k^2 + \omega^2}} \cos(\omega t - \phi) + Ce^{-kt}}\)
3 Write by heart a particular solution of this ODE?
3.1 Front
Write by heart a particular solution of this ODE?
\(\dot{x} + kx = kB \cos(\omega t)\)
3.2 Back
\({\displaystyle x_p(t) = \frac{kB}{\sqrt{k^2 + \omega^2}} \cos(\omega t - \phi)}\), where \(\phi = \operatorname{Arg}(k + i\omega)\)
4 What is the gain (or amplitude response) in this particular solution?
4.1 Front
What is the gain (or amplitude response) in this particular solution?
\(\dot{x} + kx = kB \cos(\omega t)\), where \({\displaystyle x_p(t) = \frac{kB}{\sqrt{k^2 + \omega^2}} \cos(\omega t - \phi)}\)
4.2 Back
Gain is the ratio between output amplitude and input amplitude, so
\({\displaystyle g = \frac{k}{\sqrt{k^2 + \omega^{2}}}}\)
You can write \(x_p(t) = gB\cos(\omega t - \phi)\)
5 What are the bode plots?
5.1 Front
What are the bode plots?
\(\dot{x} + kx = kB \cos(\omega t)\), where the particular solution is \({\displaystyle x_p(t) = \frac{kB}{\sqrt{k^2 + \omega^2}} \cos(\omega t - \phi)}\)
Explain how are bode plots and why are useful
5.2 Back
On the first order with constant coefficient and sinusoidal input ODE, we can get relevant information about the behaviour of the system.
We can get the gain (or amplitude response) of the system. \(g = \frac{k}{\sqrt{k^2 + \omega^{2}}}\)
Where, \(\phi\) is the phase lag, and \(g\) is the input amplitude scaled by the gain to give the output amplitude.
Fixing the coupling constant \(k\) and think how \(g\) and \(\phi\) vary as we vary \(\omega\), we can write \(g(\omega)\) and \(\phi(\omega)\)
The bode plots are graph \(g(\omega)\) and \(-\phi(\omega)\) with respect to \(\omega\) fixing the coupling constant to some values (for example, \(k = .25, .5, .75, 1, 1.25, 1.5\))
Technically, the bode plots display \(\log g(\omega)\) and \(-\phi(\omega)\) against \(\log \omega\)