How to measure the transfer function

[David83]

Hi,

In Sklar's book, he says that it is easy to measure the frequency transfer function of a LTI system by applying a sinusoidal signal as the input and observe the output on an oscilloscope. In particular, he says that if the input signal is:

\[x(t)=A\cos\left(2\pi f_0 t\right) \]

and passed through a LTI system, the output will be in the form:

\[y(t)=A|H(f_0)|\cos\left(2\pi f_0t+\theta(f_0)\right)\]

where:

\[H(f)=|H(f)|e^{j\theta(f)}\]

is the transfer function of the system.

How is that? I mean, the output of a LTI system in the time domain is the convolution between the input signal and the impulse response of the system, and in the frequency domain is the product of the Fourier transform of the input signal and the frequency transfer function. The last equation shows a multiplication between the frequency response and the signal in the time domain. How?!!

Sorry, the Latex did not work

 

[digi001]

what is a "frequency transfer function"?

frequency response and the transfer function are 2 different things.

 

[zorro]

Hi David,

A fundamental property of the complex exponential functions is that if you apply a waveform x(t)=exp(s*t) to a LTI system, where "s" is any complex number, then the output has the same form but multiplied by a complex constant that depends of "s".
Let's call H(s) that function of "s".
Then; y(t)=H(s)*exp(s*t) .
With this point of view, H(s) is the definition of the transfer function in Laplace domain.
From that, using the formula for the cos(wt)=[exp(wt)+exp(-wt)]/2, you can derive the property you mention.
Regards

Z

 

[David83]

what is a "frequency transfer function"?

frequency response and the transfer function are 2 different things.

According to Sklar's book, frequency response or frequency transfer function means the same thing.

---------- Post added at 18:29 ---------- Previous post was at 18:27 ----------

Hi David,

A fundamental property of the complex exponential functions is that if you apply a waveform x(t)=exp(s*t) to a LTI system, where "s" is any complex number, then the output has the same form but multiplied by a complex constant that depends of "s".
Let's call H(s) that function of "s".
Then; y(t)=H(s)*exp(s*t) .
With this point of view, H(s) is the definition of the transfer function in Laplace domain.
From that, using the formula for the cos(wt)=[exp(wt)+exp(-wt)]/2, you can derive the property you mention.
Regards

Z

Yes, thank. Also put \[x(t)=Re{e^{j2\pi ft}}\] will work fine.

---------- Post added at 19:00 ---------- Previous post was at 18:29 ----------

Hi zorro,

When I tried your appraoch I got:

[0.5 H(f)+0.5 H(-f)]e^{j2\pi ft}?? is H(f) an even function of f?!!

 

[LvW]

My recommendation: Choose the right forum (ANALOG) - perhaps you will get there a correct answer.
David83, are you really satisfied with an equation like this: y(t)=H(s)*exp(s*t) ?
(left side: Time function; right side: complex frequency variable).

 

[David83]

where is the problem? If you apply a complex exponential function to a system, you will get the frequency response of the system multiplied by the input. This is simply proven mathematically. Now, I am not sure about the s-domain, but this is true for the f-domain.

 

[LvW]

where is the problem? If you apply a complex exponential function to a system, you will get the frequency response of the system multiplied by the input. This is simply proven mathematically. Now, I am not sure about the s-domain, but this is true for the f-domain.

Where the problem is? Do you think you simply can multiply a function in the frequency domain (H(s) with an input that is in the time domain?
The definition of the transfer function H(s) is as follows:

H(s) is the Laplace transform of the systems impulse response - this is equivalent to the Laplace transform of the step response multiplied by "s".
H(s) cannot measured directly because it has no physical meaning. If you replace the variable "s" bei "jw" (in words: develop H(s) along the imaginary axis in the s-plane) you get the frequency response of the system F(jw).
The physical meaning of F(jw) is that it transfers the spectrum of the input signal into the output spectrum.
F8jw) has a magnitude and a phase function - and both can me measured with a swept input sinusoidal.

In summary: H(s) is a pure theoretical tool that has many advantages (poles and zeros) and it has a direct relation to the characteristic equation of the systems diff. equation (in the time domain) - and it can be transferred to the frequency response via s replacement by jw.
That's all i can say. Perhaps it helps.

 

[zorro]

Hi LvW,

In the equation y(t)=H(s)*exp(s*t), both left and right sides are time functions for any value of s.

Please read with attention what I wrote in post #3:

if you apply a waveform x(t)=exp(s*t) to a LTI system, where "s" is any complex number, then the output has the same form but multiplied by a complex constant that depends of "s".
Let's call H(s) that function of "s".
Then; y(t)=H(s)*exp(s*t) .

I can write:

For any complex constant s: if x(t)=exp(s*t) then y(t)=H(s)*x(t)


I don't agree with you in another aspect: H(s) has a physical meaning and in some cases it can be measured.

The meaning is the above: it is the "gain factor" in steady-state when the input is k*exp(s*t). This gives a way for measuring H(s) for some value of s: apply the exp at the input and look at the ratio y(t)/x(t) in steady state.

Of course, to apply in practice exponential functions at x(t) and measuring y(t) for calculate H(s) as y(t)/x(t) has some problems:
If s has a real part greater than 0 it increases with time and some problem will occur.
If s has a real part less than 0 it decreases with time and will become too small.
But some points are of practical interest beacuse they are of constant amplitude:

For s=0, exp(s*t) is 1: you are applying d.c. You look at y(t) in steady-state and you see a d.c. that is H(0) times the input. In fact, H(0) is the d.c. gain.

For s=jw, exp(s*t) is cos(wt)+j*sin(wt) :
If your system can manage complex inputs and outputs (as many digital systems do), you look at y(t) and you will see H(jw) times the input. The modulus |H(jw)| is the quotient of the amplitudes and arg(H(jw)) is the angle between phasors in steady state.
If your system is real, i.e. the coefficients of the differential equation are real and only admits real inputs and outputs (as analog systems usually do), you can not apply exp(jw*t) but cos(wt). This contains both jw and -jw components. Because of the symmetry of H(jw), you see a cosine whose amplitude and phase are the modulus and argument of H(jw).


Of course, the definition of H(s) based in the impulse response is, after all, equivalent.
But this point of view is correct, and helps to see and understand some facts. For example, that a zero corresponds to a value of s for which the forced output is null even when there is an excitation applied, or that a pole corresponds to a value of s for which there can be nonzero output even with null input...
Or, in the present case, this point of view gives the answer David83 were looking for.

Regards

Z

 

[LvW]

Hi Zorro, thanks to your long reply.
I think it is a rather interesting and important discussion about the meaning of H(s) - if s is complex.
And most important: To avoid misunderstandings regarding similar functions because the wording sometimes is not clear enough.
We should remember that David83 has asked how to measure the "Frequency transfer function".
For my opinion, this is a function like A(s) with s=jw. More than that, I am convinced that's the only transfer property you can measure in reality.
Now comes the so called transfer function H(s) - with s=a+jw - into the play.
In this respect i must disagree with your opinion:

Zorro: In the equation y(t)=H(s)*exp(s*t), both left and right sides are time functions for any value of s.
Please read with attention what I wrote in post #3:
If you apply a waveform x(t)=exp(s*t) to a LTI system, where "s" is any complex number, then the output has the same form but multiplied by a complex constant that depends of "s".

At first sight this is only a claim without proof, right?

Zorro: Let's call H(s) that function of "s".
Then; y(t)=H(s)*exp(s*t) .
I can write:
For any complex constant s: if x(t)=exp(s*t) then y(t)=H(s)*x(t)

I am convinced that this is true only if you replace H(s) bei A(jw).
For my opinion, this is evident by considering the following:
For a sinusoidal input x(t) the output y(t) of an LTI system has the same frequency as x(t) (steady state conditions) – however amplitude and phase are altered depending on the input frequency. That’s a known fact without doubt, right?
This allows to define the complex transfer factor A(jw) – and it can be shown that A(jw) is identical to the ratio of the input-output frequency spectrum. Also nothing new!
Two cases are important: If the input is x(t)=exp(s*t) we have a constant sinusoidal wave for s=jw
and we have a rising amplitude if s itself is complex: s=a+jw.

For the 2nd case, I think it is evident that the output y(t) will show the same frequency and the same rising factor a.
Now, if in the equation under discussion the input x(t) is multiplied by H(s) - with s being the same complex number as in x(t) – the rising factor a does appear not only in the signal but also in the complex function H(s) that is multiplied by x(t).
This means: The transfer properties of the LTI network depend on the input signal x(t): A clear contradiction to the LTI assumption.
Therefore: In your equation H(s) has to be replaced by A(jw).

Zorro: I don't agree with you in another aspect: H(s) has a physical meaning and in some cases it can be measured.
The meaning is the above: it is the "gain factor" in steady-state when the input is k*exp(s*t). This gives a way for measuring H(s) for some value of s: apply the exp at the input and look at the ratio y(t)/x(t) in steady state.

See my argumentation above. H(s) with s complex does not relate both time functions x(t)=exp(st) and y(t) .


Best regards
LvW

Added: One additional point: If H(s) with s=complex could be measured it also should be possible to find H(s) bei circuit simulation (transient analysis) with x(t)=rising sinusoidal wave. I doubt if this works.

 

[FvM]

It's usually assumed, that H(s) can be measured by applying either a step or delta pulse to the circuit. For linear circuits, it will at least reveal H(s) with acceptable accuracy.

For any complex constant s: if x(t)=exp(s*t) then y(t)=H(s)*x(t)

I agree, that the expression doesn't apply for signals with time varying amplitude (a ≠ 0), because the time constant a of output signal will be generally different from the input signal. A resonant circuit typically changes the signal envelope.

 

[LvW]

Hi Zorro,

in the meantime I have convinced myself that your definition as quoted below is correct:

A fundamental property of the complex exponential functions is that if you apply a waveform x(t)=exp(s*t) to a LTI system, where "s" is any complex number, then the output has the same form but multiplied by a complex constant that depends of "s".
Let's call H(s) that function of "s".
Then; y(t)=H(s)*exp(s*t) .

Here, H(s) is a constant - but dependent on a fixed complex s - that in this context is called "eigenvalue" because it relates the input x(t) with the output y(t) for one single case only: If x(t)=exp(s*t).
Insofar you are right and I have to correct myself.
For other - more realistic - input signals than exp(s*t) this factor H(s) is considered as the system function that equals the ratio between the Laplace transforms of output and input.

However, coming back to the original question from David83, I think the above explanation does not help to much.
He has asked "how to measure the transfer function" - and I think the only realistic methods are with the help of an oscilloscop or of a network analyzer, in both cases based on swept frequencies.

Regards
LvW

 

[zorro]

Hi again!

I'm sorry; I was too busy these last days, so I could not answer before now.
I would like to justify some statements I wrote in previous posts.


Imagine a LTI system for which the relationship between x(t) and y(t) is described by an ordinary linear differential equation with constant coefficients. It is known from the theory of differential equatons that the general solution for y(t) can be expressed by the sum of two parts, i.e.:

a) the solution for x(t)=0 [i.e. for the homogeneous equation]. It is the transient response, and it is a superposition of the "natural modes" of the system, each mode multiplied by some arbitrary constant.

b) a solution for the particular x(t) applied. If it does not contain any natural mode, it is called the forced or steady-state response.


Let me explain with an example.
Suppose a system with lumped elements, input x(t) and output y(t), described by the following linear differential equation:

a₂*y''(t) + a₁*y'(t) + a₀*y(t) = b₁*x'(t) + b₀*x(t)


a) Any "natural response" is the solution of the homogeneous equation

a₂*y''(t) + a₁*y'(t) + a₀*y(t) = 0

For this kind of linear equation, the solution is a sum of "natural modes" (as many as the order of the equation in y), that have the form k*exp(s_i*t). In the current case (second order) there are two modes (s₁ and s₂) that are the roots of the characteristic equation, i.e.:

a₂*s² + a₁*s + a₀ = 0.

(Of course, these are the poles of the circuit. Each natural mode is associated with a pole of the transfer function.)
The two constants that multiply the natural modes in the solution for a given problem can be determined by two conditions (frequently called "initial conditions").


b) Now we will find a particular solution (forced response) for the non-homogeneous equation for the special case of exponential input.
Suppose that we have an input x(t)=exp(s*t) where s is some complex value. In the general case t is ranging from -∞ to +∞, although any real interval can be considered. Frequently, 0 to ∞ is assumed. Te advantage of taking -∞<t<∞ is that no natural modes are excited by the transient at t=0.
For this exponential case, one particular solution of the differential equation (the forced response) is y(t) with the same shape of x(t), i.e. y(t)=H*exp(s*t) where H is some complex constant. (The theory of differential equtions teaches this, but alternatively, instead of invoke the theory we could assume that shape and later we would find that this was the right guess.)
In this case, the derivatives are: y'(t)=s*H*exp(s*t) and y''(t)=s²*H*exp(s*t).
Replacing in the differential equation, we get

a₂*s²*H*exp(s*t) + a₁*s*H*exp(s*t) + a₀*H*exp(s*t) = b₁*s*exp(s*t) + b₀*exp(s*t)

Then:

H = ( b₁*s + b₀ ) / ( a₂*s² + a₁*s + a₀ )

Remember that s can be any complex value. As the "gain" H depends of the value of s, we shall write

H(s) = ( b₁*s + b₀ ) / ( a₂*s² + a₁*s + a₀ ) for any complex s

This is our well-known transfer function.
We arrived at it solving the differential equation of the system for a complex exponential input. The result is that H(s) is the "gain" (output/input) for the forced response of the system when the exponential input is applied.

As LvW pointed out, H(s) is, for each value of s, an eigenvalue. And the complex expontial functions are the "eigenfunctions" of the linear, time-invariant transformation described by the differential equation.

I hope it is clear.
Best regards

Z

 

[LvW]

Hi Zorro,

as everybody can see, you have invested a lot of time to explain one fundamental aspect of the the relation between time and frequency domain. Thank you.
However, one short comment: At the end of part (a) you state:

(Of course, these are the poles of the circuit. Each natural mode is associated with a pole of the transfer function.)

At this point of your contribution, somebody may ask: Why?
Therefore, this sentence better should appear in part (b) after H(s) was derived.
Only now it is clear that the zeros of the denominator - identical to the poles of H(s) - are also identical to the roots of the characteristic equation that results from the differential equation in the time domain.
That is the answer to a question that was rised several times (also in this forum): What is the meaning of poles?
__________
Regards
LvW