transfer function

[meowth08]

We were given a simple circuit and we were tasked to obtain the transfer function V[SUB]out[/SUB]/V[SUB]in[/SUB]. First, I converted the given impedances to their laplace impedances. Then I simplified the circuit using continuous voltage division and I have arrived to this equation:

V[SUB]out[/SUB]/V[SUB]in[/SUB]=2/(s[SUP]2[/SUP]+3s+4)

I'm stacked on this part. I don't know how to convert the transfer function to its time domain equivalent. The roots of the denominator are imaginary. How would I deal with this? Please educate me. Thank you in advance.

 

[simonbramble]

surely to get this far you have come across s = jw. This converts from the Laplace domain to the time (or frequency) domain. Can you not just substitute jw for s. For the roots, you can determine from

[-b +/- root(b^2 - 4ac)]/2a

and get a value for s which you then translate back to jw

 

[meowth08]

roots:

-3 +/- root 9 - (4)(1)(4)
-3 +/- root (-7)

1st root: -3 + root (-7) = -3 + j7
2nd root: -3 - root (-7) = -3 - j7

I have already the roots.

Vout/Vin = 2/(s2+3s+4) = [K/(s-3+j7)] + [K*/(s-3-j7)]

Then,

K*=(2)(s-3+j7)/[(s-3+j7)(s-3-j7)] with s = 3-j7

K*=2/2j7 don't know what to do next.

 

[simonbramble]

the roots of the equation are when the denominator is equal to zero. To find the roots, you find the values, s=... as above. Replace s with jw and that gives you w, which is angular frequency. w = 2 x pi x f, so from here you can work out frequency

 

[MrAl]

We were given a simple circuit and we were tasked to obtain the transfer function V[SUB]out[/SUB]/V[SUB]in[/SUB]. First, I converted the given impedances to their laplace impedances. Then I simplified the circuit using continuous voltage division and I have arrived to this equation:

V[SUB]out[/SUB]/V[SUB]in[/SUB]=2/(s[SUP]2[/SUP]+3s+4)

I'm stacked on this part. I don't know how to convert the transfer function to its time domain equivalent. The roots of the denominator are imaginary. How would I deal with this? Please educate me. Thank you in advance.

Hi,

First off, you need to check your roots as they are not correct.

The procedure would go as follows...(simpler example to follow)...

Solve for the roots of the denominator (check your work here), there are two roots call them r1 and r2.
Form two new expressions x1 and x2 such that:
x1=s-r1
x2=s-r2

Equate the original transfer function to the partial fraction expansion:
2/(s^2+3*s+4)=A/x1+B/x2

Multiply both sides by x1*x2, simplify.
Equate powers of s to form two equations in the two unknowns A and B.
Note A and B may come out complex.

The new expanded equation will be:
A/x1+B/x2

Take the Inverse Laplace Transform of these two terms (this will be easy because they are just two exponentials [see far below]).

Use Euler Identities to convert the exponential forms into sin and cos terms (may only contain sin or just cos terms, or no sin and cos terms if the roots are not complex).

Example:

Lets do 2/(s^2+3*s+3.25). This example has simpler roots so it's more clear to see what is going on.

The denominator is:
s^2+3*s+3.25

the two roots are:
r1=-1.5+1.0*i
r2=-1.5-1.0*i

The two new expressions are:
x1=s-1.0*i+1.5
x2=s+1.0*i+1.5

Equate the original equation to the new expressions using one new unknown variable for each new expression:
2/(s^2+3*s+3.25)=A/x1+B/x2
so
2/(s^2+3*s+3.25)=A/(s-1.0*i+1.5)+B/(s+1.0*i+1.5)

Multiply both sides by x1, then multiply both sides by x2, then simplify. We get:
4=(2*s-2*i+3)*B+(2*s+2*i+3)*A

Note that this is just the denominator removed from the left side and the right side multiplied by x1 and by x2 and both sides by 2.

Expand that out we get:
4=2*s*B-2*i*B+3*B+2*s*A+2*i*A+3*A

Now we'll equate like powers of s by first arranging terms in a single column:
4
=
2*s*B
-2*i*B
+3*B
+2*s*A
+2*i*A
+3*A

now make each line contain only one power of s for each line by combining lines with like powers of s:
4
=
2*s*B+2*s*A
-2*i*B+3*B+2*i*A+3*A

(note the last line has power of s: s^0).

Now group into two equations noting that the coefficient for s on the left is zero:

0=2*s*B+2*s*A
4=-2*i*B+3*B+2*i*A+3*A


Evaluate with s=1 we get:

0=2*B+2*A
4=-2*i*B+3*B+2*i*A+3*A


Solving these two simultaneously, we get:
A=-i
B=i

So the new transfer function is:
A/x1+B/x2=-i/(s-1.0*i+1.5)+i/(s+1.0*i+1.5)

Using the Inverse Laplace Transform for these two terms, we get:
-i*e^(-(1.5-1.0*i)*t)+i*e^(-(1.5+1.0*i)*t)

Using Euler Identities, we convert this to sin and/or cos terms and end up with:
f(t)=2*e^(-1.5*t)*sin(t)

and in terms of Vin and Vout:
Vout/Vin=2*e^(-1.5*t)*sin(t)

A small note about the Inverse Laplace Transform being used here...

Looking back, the Inverse Laplace Transform is easy to use here because for a general frequency a/x term:
a/(s-Imag*i+Real)

the Inverse Laplace is always an exponential:
a*e^-(t*(Real-i*Imag))

no matter how many terms we have. The last step would be to convert these exponential terms into sin and/or cos terms, and that would give us the final time equation.

If the Imag part is positive it's the same except for the sign of the imaginary part of rhe exponent:
a/(s+Imag*i+Real) => a*e^-(t*(Real+i*Imag))

 

[meowth08]

I already finished this homework and my answer is different to Mr AI's approach. However, I'll copy this one and try resolving the problem so would not repeat the same mistake I did. I'll post a reply as long as I finish working on it. Thank you Mr. AI.

 

[meowth08]

I already resolved the problem. I was wrong when I said that Mr. AI's approach was different from my answer. I would say that Mr AI's detailed example explained to me the parts of my homework I didn't understand before because I only patterned my solution before to a similar problem in the internet not knowing that what I was using was Euler's theorem. Thanks again sir.

 

[MrAl]

Hi again,


You're welcome. BTW my screen name is MrAl (last letter is lower case English "L" not upper case "I") but i get that from time to time because the font used on the internet has characters that often look the same and rely on the context to tell them apart. With a name like MrAl it's harder to figure out the context so i have to point it out now and then.

In some courses they teach how to go right from the equation to the transform exponential terms. I happen to think it's good to know the steps in between too rather than a more elaborate grouping theme to get to the end result.