### Welcome to EDAboard.com

#### happyone21

##### Newbie

1.Find the frequency response of this circuit.

2.Explain which filter this circuit is.

3.Find the impulse response in this circuit.

4. When x(t)=e^(-3t) u(t) is applied as input, find the output y(t).

#### KlausST

##### Super Moderator
Staff member
Hi,

I don't call it a filter at all.

A filter has R and C in series ... which isn't the case here.

Klaus

#### FvM

##### Super Moderator
Staff member
I presume x(t) is a current source. It's a first order low-pass.

#### andre_teprom

##### Super Moderator
Staff member
Have a look on Thevenin and Norton equivalent circuits.
The leftmost resistor and current source in parallel could be replaced by an equivalent resistor and voltage source in series.

#### barry

You don't need Rs and Cs in series to have a filter. This is a low pass filter. At low frequency, the capacitor is essentially infinite resistance. At high frequency the capacitor is essentially a short.

If you use Laplace transforms for this circuit (assuming x(t) is current and y(t) is voltage), you'll have: y(s) = x(s)*(1/R*1/sC)/(1/R+1/sC)

Last edited:

#### KlausST

##### Super Moderator
Staff member
Hi,

I can´t see where it is stated that x(t) is a current.
And if x(t) is a current, then I expect y(t) to be a current, too.

Guessing x(t) is a current and guessing y(t) is a voltage ( yes I know it´s maybe the only way to give the circuit a meaningful function)
... is this the target of the homework?

***
You don't need Rs and Cs in series to have a filter. This is a low pass filter. At low frequency, the capacitor is essentially infinite resistance. At high frequency the capacitor is essentially a short.
I can´t agree.
At low frequency there is finite impedance. Only at DC there is infinite impedance.
At high frequency there still is impedance. Only at infinte frequency there is a short.
(both is not the case with x(t))

And guessing that x(t) is a current... means there is a current source, which has infinite source impedance.

***
So now one could say we don´t have to calculate with ideal parts ... but if not´it´s getting even more complicated...

***

In the end I´d say
* either it´s a trick question
* or there is a lack of information.

Klaus

#### barry

Wouldn’t you agree at low frequency a capacitor’s impedance is really, really high? And at high frequency it’s really, really low? You don’t need zero or infinite impedance to understand the frequency response of this circuit which:

1) shows x(t) as a current source, and
2) shows y(t) obviously as a voltage.

#### KlausST

##### Super Moderator
Staff member
Hi,

For sure I agree with high and low impedance...

But what I really doubt is, that it's the target of the exercise to find out that x(t) needs to be current and y(t) needs to be voltage to get useful function.

For sure people working in electronics design will find out, but I'd say at least 50% of my electronics teachers (without practical experience) are not able to find this out.

Klaus

Staff member

#### KlausST

##### Super Moderator
Staff member
Hi,

This really makes sense. Thanks a lot.
This is what I have missed.

Klaus

#### Easy peasy

x(t) is a current source which is driven by u(t)

the OP needs to find y(t) as a function of u(t)

Replies
1
Views
1K
Replies
0
Views
860
Replies
3
Views
902
Replies
6
Views
2K
Replies
0
Views
943