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Sinusoidal Steady State Response

Cause and effect for linear systems are related by linear differential equations. note that for the exponential function $ y=Ke^{ax}$,

$\displaystyle \frac{dy}{dx}=Kae^{ax}$    
and $\displaystyle \frac{d^y}{dx^2}=K_2e^{ax}$    

where $ K_2=a^2K$ is also a constant. Similar equations hold for $ n^{th}$ derivatives also.
Figure 8.1: An Linear System

Consider a function $ f(x)$. If on operation by a system, the function $ f(x)$ is only multiplied by a constant $ k$, the function is said to be the eigenfunction of the given system, as shown in 8.1.
We know that:

$\displaystyle sin(x)=\frac{e^{jax}+e^{-jax}}{2j}$ (8.1)

Figure 8.2: Superposition

Consider a linear system, as in 8.2. The functions $ f_1(x)$ and $ f_2(x)$ are eigen functions of the linear system. If sum of these two functions is input to the system, the output can be predicted easily, by superposition theorm. Thus knowing the output to eigenfunctions helps us in predicting output of several other functions.

Now we look at how output to sinusoidal excitations of linear circuits can be determined. Consider $ sin(\omega
t)=\frac{e^{jax}-e^{-jax}}{2j}$ or $ cos(\omega

Note that $ \frac{d(sin(\omega t))}{dt}=\omega cos(\omega t)=\omega
sin(\omega t+\frac{\pi}{2})$. Thus, the phase and associated constant changes when a sinusoid is passed through a differentiator.

Similarly, $ \frac{d(cos(\omega t))}{dt}=-\omega sin(\omega t)=\omega
cos(\omega t+\frac{\pi}{2})$.

Figure 8.3: RLC circuit

Now consider the RLC circuit shown in 8.3.

$\displaystyle v(t)=Ri(t)+L\frac{di}{dt}+\int_{-\infty}^ti(t)dt$    
If i(t) is sinusoidal, say cos(wt), then    
$\displaystyle v(t)=cos(wt)-w\;sin(wt)+\frac{1}{w}sin(wt)$    
$\displaystyle =cos(wt)-(w-\frac{1}{w})cos(wt)$    
$\displaystyle =\sqrt{1+(w-\frac{1}{w})}\left( \frac{1}{\sqrt{1^2+(w-\frac{1}{w})^2}} - \frac{w-\frac{1}{w}}{\sqrt{1^2+(w-\frac{1}{w})^2}sin(wt)} \right)$    
$\displaystyle =\sqrt{1+(w-\frac{1}{w})}cos(wt+\theta)\;\;\;$where, $\displaystyle \; \theta =tan^-1(w-\frac{1}{w})$    

the last equality follows using $ cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$

$ i(t)=cos(wt)$
$ v(t)=A\;cos(wt+\theta)$, as shown in 8.4.

Figure 8.4:

Analysis can be done simply using sin and cos terms.

But can't be further simplified using imaginary quantities
$ v(t)=A\;cos(wt+\theta)+j\;A\;sin(wt+\theta)$

Figure 8.5:

As in the diagram shown in 8.5, $ R=\sqrt{A^2sin^2(wt+\theta)+A^2cos^2(wt+\theta)}=A$ In figure 8.6, the complex values shown are rotating with time. The actual value at any time is the projection on the real axis.

Figure 8.6: Values rotating in the complex plane with time

Note that $ v(t)$ and $ i(t)$ have constant separation with each other. All entities in general will have same relative separation.

Figure 8.7: Phasors

Consider figure 8.7. Let us rotate the frame of reference (or axis) also with speed $ w$ rad/sec. Then the angles of $ v(t)$ and $ i(t)$ with respect to the axis become constants. The figure 8.7 is a phasor diagram, showing current and voltage phasors, and the phase difference between the two.

We show the application of phasors in circuit analysis by the circuit shown in figure 8.8, a simple inductor circuit excited by a sinusoidal voltage source.

Figure 8.8: Inductor circuit

$\displaystyle v(t)=cos(wt)$    
$\displaystyle v(t)=L\frac{di}{dt}$    
$\displaystyle i(t)=\frac{1}{L}\int v(t)dt=\frac{1}{L}sin(wt)$    
$\displaystyle =\frac{1}{L}cos(wt-\pi/2)$    

The associated phasor diagram is shown in figure 8.9. It can be seen that the phase difference is $ \pi/2$ radians. Voltage leads the current by $ \pi/2$ radians.
Figure 8.9: Phasor diagram for inductor excitation

Figure 8.10: Resistance

Similarly, consider a resistance excited by a sinusoidal votage source, as shown in figure 8.10. Here, $ v(t)=cos(wt)$, and $ i(t)=\frac{v(t)}{R}=\frac{cos(wt)}{R}$ is in phase with the voltage. The same is shown in the phasor diagram 8.11.

Figure 8.11: Phasor for resistor excitation

Figure 8.12: Resistor and inductor in series

Now consider a resistor and an inductor in series with an AC voltage source, as in 8.12.

$\displaystyle i(t)=cos(wt)$    
$\displaystyle v_R=R\;cos(wt)$    
$\displaystyle v_L(t)=-Lw\;sin(wt)=wL\;cos(wt+\pi/2)$    
$\displaystyle v(t)=v_L(t)+v_R(t)=R\;cos(wt)+wL\;cos(wt+\pi/2)$    
$\displaystyle v(t)=Re\{R\;e^{wt}+wL\;e^{wt+\pi/2}\}$    
$\displaystyle =Re\{(R\;cos(wt)+wL\;cos(wt+\pi/2))+j(Rsin(wt)+wl\;sin(wt+\pi/2))\}$    
$\displaystyle =Re\{(R\;cos(wt)-wL\;sin(wt))+j(Rsin(wt)+wl\;cos(wt))\}$    
$\displaystyle =Re\{\sqrt{R^2+w^2+L^2}(cos(wt+\theta))+j\sqrt{R^2+w^2+L^2}sin(wt+\theta)\}$    
where $\displaystyle \theta=tan^{-1}(\frac{wL}{R})$    

The resulting phasor diagram is plotted in figure 8.13.

Figure 8.13: Phasor

This gives an inkling to a general result: phasors can be added/subtracted just like vectors. Resulting magnitude and phase would come out to be the same. See the hint below.

In phasor terms: Voltage across inductor: $ V_L=jIwL$, Voltage across resistor$ =V_R=IR$. Now adding the two vectors, $ V_{L,R}=V_R+V_L=IR+jIwL=I\sqrt{R^2+w^2L^2}\;tan^{-1}(\frac{wL}{R})$

Compairing with Ohm's law, $ V=IR$, the previous equation $ V=I(R+jwL)$, the complex term $ R+jwL$ can be taken to be similar to resitance. This is called impedance. Inverse of impedance is called admittance, complex analog of conductance. In the above circuit,

$\displaystyle I=V\frac{1}{R+jwL}(\frac{R-jwL}{R-jwL})=\frac{v(R-jwL)}{R^2+w^2L^2}$    
$\displaystyle \Rightarrow I=V\left(\frac{R-jwL}{R^2+w^2L^2} \right)$    

Similar to the above analysis, we now work with the capacitor. See figure 8.14.

$\displaystyle i_c=cos(wt)$    
$\displaystyle v_c=\frac{1}{C}\int i_c dt=\frac{1}{wC}sin(wt)$    
$\displaystyle =\frac{1}{wC}cos(wt-\pi/2)$    

Thus, $ v_c$ lags by $ \pi/2$ w.r.t. $ i_c$, as shown in phasor diagram 8.15. The phasors are typically written in capital letters, whereas their continuous time counterparts in small letters. In phasor diagram, $ V_c=\frac{-j}{wC}I_c=\frac{1}{jwC}I_c$. Thus, impedance is $ \frac{1}{jwC}$.
Figure 8.14:
Figure 8.15:

Figure 8.16:

Now we analyse circuit shown in in figure 8.16, $ i(t)=I_Rsin(wt+\theta)$. We wish to calculate $ I_R$ and $ \theta$.

$\displaystyle I_1=I_1\angle\theta_1=I_{1x}+jI_{1y}$   (phasor in rectangular coordinates x and y)    
$\displaystyle I_2=I_2\angle\theta_2=I_{2x}+jI_{2y}$    
for first loop: $\displaystyle I_1jwL+(I_1-I_2)\frac{1}{jwC}=V$ ...(*)    
for second loop: $\displaystyle (I_2-I_1)\frac{1}{jwC}+I_2(R_1+R_2)=0$ ...(**)    
From (**): $\displaystyle \frac{-I_1}{jwC}+I_2\left(\frac{1}{jwC}+R_1+R_2 \right)$   From (*) :$\displaystyle I_1(jwL-\frac{1}{jwC})-I_2\frac{1}{jwC}=V$    

Solving these two for $ I_1$ and $ I_2$, we get:

$\displaystyle I_2=\frac{V}{jwL-(R_1+R_2)w^2LC-\frac{2}{jwC}-(R_1+R_2)}$      
which after rationalization, gives:       
$\displaystyle I_2=\frac{VwL}{\sqrt{(2+w^2LC)+(wC(R_1+R_2)(1+w^2LC))^2}}\angle{tan^{-1}(\frac{wC(R_1+R_2)(1+w^2LC)}{2+w^2LC})}$      

For sinusoidal forcing functions, we can use the same techniques, but with complex variables

A sinusoid $ sin(wt)$ passed through a linear system with transfer function $ H(w)$, the output would be $ \vert H(w_1)\vert sin(w_1t+\angle{H(w_1)})$.

Thus, for a sum of sinusoids of different frequencies, using superposition principle, the output for $ x(t)=a_1sin(w_1t)+a_2sin(w_2t)+\ldots$ would be:

$\displaystyle y(t)=a_1\vert H(w_1)\vert sin(w_1t+\angle{H(w_1)}+a_2\vert H(w_2)\vert sin(w_2t+\angle{H(w_2)})+\ldots$ (8.2)

next up previous contents
Next: Power Supply Up: Introduction to Electronics Previous: R-C Circuits   Contents
ynsingh 2007-07-25