Laplace transform in electrical engineering

In summary, the conversation discusses the use of Laplace transform in solving circuits and the connection to phasor techniques. It is explained that the Laplace transform can be used to find the steady state response of a stable system to a t-domain input. The process of finding the steady state response is also discussed, including the relevance of only terms with purely imaginary poles and the calculation of the steady state response in the s-domain. The conversation ends with a brief explanation of how to find the steady state response using the Laplace transform.
  • #1
Bassalisk
947
2
Hello,

We recently learned in mathematics the Laplace transform. But we didn't actually use it to solve circuits yet.

I took things at my own hand and tried to figure out what's the point of Laplace transform. I figured out all that frequency and time stuff. But here is the thing.

If you Laplace transform an equation that describes let's say capacitor(in time domain), you get

I=V*wC-I0*C right?

I see resemblance with impedance and other phasor techniques we used to solve circuits. But where is that imaginary part?

From transform up top. Xc=1/wC. But its missing that "-j" part. How did people get to the equation Xc=-j*1/wC?
 
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  • #2
I don't know what system you're trying to solve but if you just look at the Laplace transform for a single capacitor

You have [itex] i(t) = C \frac{dv}{dt} <==> i(s) = C(s*V(s) - V(0)) ==> V(s) = \frac{i(s)}{sC} + \frac{V(0)}{s} [/itex]

Notice that i(s) and V(s) both have an s dependence. s is defined as a+jb so there is indeed a complex component.
 
  • #3
Feldoh said:
I don't know what system you're trying to solve but if you just look at the Laplace transform for a single capacitor

You have [itex] i(t) = C \frac{dv}{dt} <==> i(s) = C(s*V(s) - V(0)) ==> V(s) = \frac{i(s)}{sC} + \frac{V(0)}{s} [/itex]

Notice that i(s) and V(s) both have an s dependence. s is defined as a+jb so there is indeed a complex component.

oooh i thought s was omega :D Thank you !
 
  • #4
If H(s) is a transfer function of a stable system and the t-domain input is [itex]cos(\omega t + \theta)[/itex], then the steady state output is [itex]\left|H(j\omega)\right|cos(\omega t + \theta + \angle H(j\omega))[/itex]
This means if [itex]Z(s) = V(s)/I(s)[/itex], we may just multiply the current phasor by [itex]Z(j\omega)[/itex] to get the voltage phasor.

assuming zero initial capacitor voltage
[itex]\frac{V(s)}{I(s)} = \frac{1}{sC}[/itex], with [itex]i(t)
= Re\left\{e^{j\omega t + \theta}

\right\}[/itex]
then
[itex]v(t) = Re\left\{\left(\frac{1}{j\omega C}\right)e^{j\omega
t + \theta}\right\}[/itex]

understanding why this is:

As H(s) is the transfer function of a stable system, the real part of all the poles of H(s) will be less than zero.

In the partial fraction expansion of the system's response to [itex]cos(\omega t)[/itex], only the terms with purely imaginary poles will be relevant to the steady state, as the other terms (with poles in the left half plane) will correspond to decaying exponentials in the t-domain.

With the Laplace transform of [itex]cos(\omega t + \theta) = cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta)[/itex] being [itex]X(s) = cos(\theta)\frac{s}{(s + j\omega)(s - j\omega)} - sin(\theta)\frac{\omega}{(s + j\omega)(s - j\omega)}[/itex], the s-domain response of the system is

[itex]H(s)X(s) = \frac{k}{s - j\omega} + \frac{k*}{s + j\omega} + N(s)[/itex]

where N(s) is the fraction expansion for all the other poles of the response. So

[itex]H(s)X(s)(s - j\omega) = k + (\frac{k*}{s + j\omega} + N(s))(s - j\omega)[/itex]

(s - j\omega) in the numerator and denominator of the left hand side cancel after substituting in the expression for X(s).

[itex]H(s)\frac{cos(\theta)s - sin(\theta)\omega}{s + j\omega} = k + (\frac{k*}{s + j\omega} + N(s))(s - j\omega)[/itex]

taking the limit of both sides as s goes to [itex]j\omega[/itex]

[itex]\lim_{s \to j\omega}H(s)\frac{(cos(\theta)s - sin(\theta)\omega)}{(s + j\omega) } = k + (\frac{k*}{s + j\omega} + N(s))(0)[/itex]

so

[itex]k = H(j\omega)\frac{(j\omega cos(\theta) - \omega sin(\theta))}{(2j\omega) } = \frac{H(j\omega)}{2}(cos(\theta) - (-jsin(\theta)) = \frac{e^{j\theta}H(j\omega)}{2} [/itex]

the s-domain steady state response is

[itex]\frac{k}{s - j\omega} + \frac{k*}{s + j\omega}[/itex]

Which in the t domain becomes

[itex]\left|k\right|e^{\angle k}e^{j\omega t} + \left|k\right|e^{-\angle k}e^{-j\omega t} = 2 \left|k\right|cos(\omega t + \angle k)[/itex]

substituting for k:
[itex]\left|k\right| = \frac{\left|H(j\omega)\right|}{2}[/itex]
[itex]\angle k = \theta + \angle H(j\omega)[/itex]

[itex]2 \left|k\right|cos(\omega t + \angle k) = \left|H(j\omega)\right|cos(\omega t + \theta + \angle H(j\omega))[/itex]
 
Last edited:
  • #5
MisterX said:
If H(s) is a transfer function of a stable system and the t-domain input is [itex]cos(\omega t + \theta)[/itex], then the steady state output is [itex]\left|H(j\omega)\right|cos(\omega t + \theta + \angle H(j\omega))[/itex]
This means if [itex]Z(s) = V(s)/I(s)[/itex], we may just multiply the current phasor by [itex]Z(j\omega)[/itex] to get the voltage phasor.

assuming zero initial capacitor voltage
[itex]\frac{V(s)}{I(s)} = \frac{1}{sC}[/itex], with [itex]i(t)
= Re\left\{e^{j\omega t + \theta}

\right\}[/itex]
then
[itex]v(t) = Re\left\{\left(\frac{1}{j\omega C}\right)e^{j\omega
t + \theta}\right\}[/itex]



understanding why this is:

As H(s) is the transfer function of a stable system, the real part of all the poles of H(s) will be less than zero.

In the partial fraction expansion of the system's response to [itex]cos(\omega t)[/itex], only the terms with purely imaginary poles will be relevant to the steady state, as the other terms (with poles in the left half plane) will correspond to decaying exponentials in the t-domain.

With the Laplace transform of [itex]cos(\omega t + \theta) = cos(\omega t)cos(\theta) - sin(\omega t)sin(\theta)[/itex] being [itex]X(s) = cos(\theta)\frac{s}{(s + j\omega)(s - j\omega)} - sin(\theta)\frac{\omega}{(s + j\omega)(s - j\omega)}[/itex], the s-domain response of the system is

[itex]H(s)X(s) = \frac{k}{s - j\omega} + \frac{k*}{s + j\omega} + N(s)[/itex]

where N(s) is the fraction expansion for all the other poles of the response. So

[itex]H(s)X(s)(s - j\omega) = k + (\frac{k*}{s + j\omega} + N(s))(s - j\omega)[/itex]

(s - j\omega) in the numerator and denominator of the left hand side cancel after substituting in the expression for X(s).

[itex]H(s)\frac{cos(\theta)s - sin(\theta)\omega}{s + j\omega} = k + (\frac{k*}{s + j\omega} + N(s))(s - j\omega)[/itex]

taking the limit of both sides as s goes to [itex]j\omega[/itex]

[itex]\lim_{s \to j\omega}H(s)\frac{(cos(\theta)s - sin(\theta)\omega)}{(s + j\omega) } = k + (\frac{k*}{s + j\omega} + N(s))(0)[/itex]

so

[itex]k = H(j\omega)\frac{(j\omega cos(\theta) - \omega sin(\theta))}{(2j\omega) } = \frac{H(j\omega)}{2}(cos(\theta) - (-jsin(\theta)) = \frac{e^{j\theta}H(j\omega)}{2} [/itex]

the s-domain steady state response is

[itex]\frac{k}{s - j\omega} + \frac{k*}{s + j\omega}[/itex]

Which in the t domain becomes

[itex]\left|k\right|e^{\angle k}e^{j\omega t} + \left|k\right|e^{-\angle k}e^{-j\omega t} = 2 \left|k\right|cos(\omega t + \angle k)[/itex]

substituting for k:
[itex]\left|k\right| = \frac{\left|H(j\omega)\right|}{2}[/itex]
[itex]\angle k = \theta + \angle H(j\omega)[/itex]

[itex]2 \left|k\right|cos(\omega t + \angle k) = \left|H(j\omega)\right|cos(\omega t + \theta + \angle H(j\omega))[/itex]

Thank you I will have a look into this in more detail later.
 
  • #6
correction

MisterX said:
...[itex]i(t) = Re\left\{e^{j\omega t + \theta} \right\}[/itex]
then
[itex]v(t) = Re\left\{\left(\frac{1}{j\omega C}\right)e^{j\omega t + \theta}\right\}[/itex]

should be
[itex]i(t)
= Re\left\{e^{j(\omega t + \theta)}\right\}[/itex]
then
[itex]v(t) = Re\left\{\left(\frac{1}{j\omega C}\right)e^{j(\omega t + \theta)}\right\}[/itex]
 

Related to Laplace transform in electrical engineering

1. What is the Laplace transform and how is it used in electrical engineering?

The Laplace transform is a mathematical tool used to transform a function from the time domain to the frequency domain. This allows engineers to analyze and solve problems related to electrical circuits and systems, such as finding the response of a circuit to a specific input signal.

2. How is the Laplace transform different from the Fourier transform?

While both the Laplace transform and Fourier transform are used to analyze signals in the frequency domain, the Laplace transform also takes into account the initial conditions of a system, making it more suitable for analyzing dynamic systems in electrical engineering.

3. What is the inverse Laplace transform and how is it used in electrical engineering?

The inverse Laplace transform is used to transform a function back from the frequency domain to the time domain. This allows engineers to obtain the time-domain solution of a circuit or system by using the frequency-domain representation obtained through the Laplace transform.

4. What are some common applications of the Laplace transform in electrical engineering?

The Laplace transform is commonly used in electrical engineering to analyze and design control systems, filter circuits, and signal processing algorithms. It is also used in the analysis of stability and transient response of electric circuits and systems.

5. Are there any limitations to using the Laplace transform in electrical engineering?

While the Laplace transform is a powerful tool, it is limited to linear systems and cannot be directly applied to nonlinear systems. In addition, it requires advanced mathematical knowledge to accurately apply and interpret the results in the context of electrical engineering.

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