Evaluating Laplace Transform at a specific frequency

[hkBattousai]

I'm trying to find an expression for current at a branch in my circuit. I used Laplace transform to find that expression. The expression contains a term "V(s)" which is the supply voltage.

v(t) = 310 . cos(2π50t) V (220V, 50Hz alternating sinusoidal voltage)

V(s) = 310 . [ s / (s^2 + ω^2)], where ω is 2π50 rad/s

We substitute,
s = σ + jω, where σ=0 and ω=2π50 rad/s
That makes,
s = j2π50 rad/s

We substitute s into V(s):
V(s) = 310 . [ s / (s^2 + ω^2)]
V(s) = 310 . [ (j2π50) / ((j2π50)^2 + (2π50)^2)]
V(s) = 310 . [ (j2π50) / (-(2π50)^2 + (2π50)^2)] = 310 . [ (j2π50) / 0] = j∞
V(s) = j∞

What is wrong here, why can't I find a finite value for V(s)?

 

[Ratchit]

hkBattousai,

You don't get a finite numerical value for a Laplace transform. That is because it is in the frequency domain. If you want a finite value, transform it into the time domain by finding the inverse Laplace transform of the function, and substitute the parameters for a real time value. Why are you trying to substitute σ+jω into the Laplace transform of the sine? You already know the Laplace transform of the sine. Maybe you better look at Laplace Transform of Exp and sine .

Ratch

 

[hkBattousai]

hkBattousai,

You don't get a finite numerical value for a Laplace transform. That is because it is in the frequency domain. If you want a finite value, transform it into the time domain by finding the inverse Laplace transform of the function, and substitute the parameters for a real time value. Why are you trying to substitute σ+jω into the Laplace transform of the sine? You already know the Laplace transform of the sine. Maybe you better look at Laplace Transform of Exp and sine .

Ratch

Yeah, I got it now. You're saying that I have to convert s-domain form into time-domain form and this is the only way. Evaluating V(s) does not give any meaningful value for time-domain.

Thank you.

 

[hkBattousai]

I want to share the results of my work.

Page #1:
**broken link removed**

Page #2:
**broken link removed**

My aim is to obtain 6Vp-p sinusoidal voltage from 220V, 50Hz voltage source. I wanted to analytically see the affects of changing load resistance on load voltage and load current.

At the bottom of the second page, you can see the time domain expressions of voltage and current on load.

 

[MrAl]

Hi there,


I dont know what gave you the idea that you can not obtain the output without resorting to the time domain expression. Of course you surely can when the excitation is a sine wave as it is here.

For some reason though you seem to have approached the problem incorrectly from the start. The idea is to develop the Laplace equation for IL(s) from the given network and input, then replace every 's' with j*w, then evaluate the output amplitude and if desired also the phase angle relative to the input. That's a heck of a lot easier and faster than trying to determine the time domain response for every network.

For your circuit example with the two capacitors and one load resistor you get a transform like this:
IL(s)=Vin*(s*C1)/((C1+C2)*s*RL+1)

and checking this equation with s=j*w (at 50Hz) and RL=100 ohms and C1=C2=1000uf, i get just under 5ma peak through RL, which seems correct with Vin=1v peak.

The output amplitude (RL current) can be written in an explicit form too from IL(s) as:
IL(ampl)=Vin*w*C1/sqrt(w^2*RL^2*(C1+C2)^2+1)

You dont really need to know the phase angle for this application unless you want to write the time domain equations too and use the phase angle and the IL amplitude to double check the results, making sure to get the same answer both ways.

I think you should try this again and if you have a problem i'll go over it with you no problem.

As a side note, in many applications like this one you can eliminate C2 and just use C1 and calculate the current that way, adjusting C1 as needed. The only other thing to consider in either design is the surge current during turn on and a possible bleeder resistor for the cap(s).

 

[hkBattousai]

The circuit:
**broken link removed**



Expression for load voltage:

[latex]v_L(t)\,=\,V_m\,\frac{C_1}{C_1+C_2}\Huge(\normalsize\frac {1}{R^2(C1+C2)^2\omega^2+1}\,\Large e\normalsize\,^{(-\frac{\omega t}{R(C_1+C_2)\omega})}\,+\,\frac{R(C_1+C_2)\omega\sqrt{R^2(C_1+C_2)^2\omega^2+1}}{R^2(C_1+C_2)^2\omega^2\,+\,1}\,\cos\LARGE(\normalsize wt+arctan(-\frac{1}{R(C_1+C_2)\omega})\LARGE)\normalsize\Huge)\normalsize[/latex]

(Vm is the peak level of the sinusoidal source voltage.)



MATLAB code for circuit simulation:

% Circuit constants
C1 = 50 * 10^(-9);
C2 = 1000 * 10^(-9);
R = 2 * 10 ^ 3;
w = 2 * pi * 50;
VMAX = 220 * sqrt(2);

% Simulation time limit
TMAX = 0.1;

% Simulation time resolution
TRES = 0.0001;

for ti = 0 : 1 : (TMAX / TRES) - 1
    % tt: integer time index to use in vectors
    % t : actual time value
    t = ti * TRES;
    
    % Intermediate variables that simplifies the expression
    k = R * (C1 + C2) * w;
    k21 = (k * k) + 1;
    phi = atan(-1 / k);
    
    % Calculate voltage sample
    v_trans = VMAX * (C1 / (C1 + C2)) * (1 / k21) * exp(-w * t / k);
    v_stead = VMAX * (C1 / (C1 + C2)) * (k * sqrt(k21) / k21) * cos((w * t) + phi);
    v(ti + 1) = v_trans + v_stead;
    
    % Calculate current sample
    i(ti + 1) = v(ti + 1) / R;
end

% Index for plotting
tt = 0 : TRES : (TMAX / TRES) - 1;

% Truncate workspace variables
tt = tt(1 : TMAX / TRES);
i = i(1 : TMAX / TRES);
v = v(1 : TMAX / TRES);

% Plot current
hold off;
plot(tt, v, 'r');
hold on;
plot(tt, 1000 * i, 'b');
xlabel('Time (s)');
ylabel('Current and Voltage (mA and V)');

Voltage and current curves:
**broken link removed**
Voltage source: 220V, 50Hz, sinusiodal
ω=2π50 rad/s
C1 = 50nF
C2 = 1µF
R = 2kΩ

 

[MrAl]

Hi again,


Why do you want to go through the trouble of transforming to the time domain if you are using sinusoidal stimulation?

 

[hkBattousai]

Why do you want to go through the trouble of transforming to the time domain if you are using sinusoidal stimulation?

Just having fun by getting carried away with all those analytic work...

 

[MrAl]

Hi again,


Oh ok i see, well did you try it the other way too then?

 

[hkBattousai]

Oh ok i see, well did you try it the other way too then?

No, I didn't try steady state analysis since I found the general solution.
But next time, I will try the simpler way.

Thank you.

 

[MrAl]

Hi again,

We can do the simpler circuit with just one cap and one resistor too so you can compare. In this circuit, there is one cap C and one resistor R in series, and we take the output across the resistor R.

Starting with the component transforms:
resistor: R
cap: 1/(s*C)

Voltage across resistor:
vR=Vin*R/(R+1/(s*C))

Simplified:
vR=Vin*(s*C*R)/(s*C*R+1)

Replace each 's' with 'j*w': Vin*(j*w*C*R)/(j*w*C*R+1)

Multiply top and bottom by the conjugate of the bottom:

vR(jw)=Vin*j*w*C*R*(1-j*w*C*R)/((1-j*w*C*R)*(j*w*C*R+1))

vR(jw)=Vin*(w^2*C^2*R^2+j*w*C*R)/(w^2*C^2*R^2+1)

Set D equal to the denominator:
D=(w^2*C^2*R^2+1)

then the real part is:
real=w^2*C^2*R^2/D

and the imaginary part is:
imag=w*C*R/D

so the amplitude is:
vR(ampl)=Vin*sqrt(real^2+imag^2)

which equals:
vR(ampl)=Vin*sqrt((w^4*C^4*R^4)/(w^2*C^2*R^2+1)^2+(w^2*C^2*R^2)/(w^2*C^2*R^2+1)^2)

which after simplification becomes:
vR(ampl)=Vin*(abs(w)*abs(C)*abs(R))/sqrt(w^2*C^2*R^2+1)

and drop the 'abs' functions with w, C, and R always positive:
vR(ampl)=Vin*w*C*R/sqrt(w^2*C^2*R^2+1)

Note that we dont have to solve for the peak here either.

 

[hkBattousai]

My aim is to obtain 6Vp-p sinusoidal voltage from 220V, 50Hz voltage source. I wanted to analytically see the affects of changing load resistance on load voltage and load current.

The resistor in the circuit is actually not a resistor. It represents the rectifier, filter and the load. In other words, I will connect a bridge rectifier parallel to [LATEX]C_2[/LATEX] capacitor. Next I will use a 7805 to filter this rectified voltage, and then feed a more complex circuit with it.

In case we use a series R-C structure, what if R becomes too large (i.e.; the main circuit draws no current)? There will be too much voltage on bridge rectifier and 7805 which will damage them in no time.

Maybe using a zener diode for over-voltage protection be a solution.

 

[MrAl]

Hello again,


My example of the RC circuit was really just that, an example, just to show how quick the solution can be. Im not definitely saying that you should use that circuit without some discretion of course. In your app it may or may not be a good idea.
Yes, if R becomes large then the voltage will rise, and yes again that's where a zener diode could help perhaps. With the two cap circuit you'll also have to consider the total current draw from the main supply which may be higher than you want using two caps.

Power supplies like the one you are talking about have been made for lower current devices for some time now. A cap is used to drop some of the voltage without using power that a resistor would normally require. This relaxes the strain on the main supply quite a bit and also reduces heat in the circuit that would be caused by a hefty dropping resistor. That means the whole thing runs cool.
A good example is powering an LED from the mains supply like 120vac. If you would like details let me know.