- #1
brendan_foo
- 65
- 0
Hi there,
I'm trying to derive an expression for the transient response (for a step input of magnitude V), for a non-ideal inductor modeled in the schematic I have drawn. This non-ideal inductor includes its inductance ( ), a parasitic parallel resistance and a parasitic capacitance.
The schematic is here: http://homepage.ntlworld.com/b.preece/RLC.JPG
So my initial thoughts are:
[tex]Z_{c}(S) = \frac{1}{cS}[/tex]
[tex]Z_{l}(S) = lS[/tex]
[tex]Z_{r}(S) = R[/tex]
Right... so the parallel combination of all of these three operational impedances will be defined as:
[tex]Z_{eff} = \left[\frac{1}{Z_{r}} + \frac{1}{Z_{c}} + \frac{1}{Z_{l}}\right]^{-1}[/tex]
So using those values quoted up above, we should have :
[tex]Z_{eff} = \left[\frac{1}{R} + Cs + \frac{1}{sL}\right]^{-1}[/tex]
Which if I'm not mistaked is evaluated to:
[tex] Z_{eff} = \frac{sLR}{(LCR)s^2 + sL + R_p} [/tex]
[itex] R_p [/itex] Any subscript 'p' indictating that its a parasitic value.
Ok... Now based upon the Voltage divider equation, we can say that the transfer function is as below:
[tex] H(s) = \frac{V_{out}}{V_{in}} = \frac{Z_{eff}}{Z_{eff} + Z_{fix}} [/tex]
Now we have:
[tex]{Z_{eff} + Z_{fix} = \frac{sLR + R_{fix}((LCR)s^2 + sL + R_{p})}{(LCR)s^2 + sL + R_p}[/tex]
And from there we have the Transfer function [itex]H(s)[/itex] to be:
[tex] H(s) = \frac{sLR}{sLR + R_{fix}((LCR)s^2 + sL + R_{p}}[/tex]
Knowing that the Laplace transform of the unit step function at t = 0, multiplied by a scaling value will be:
[tex] L(u(t)) = \frac{1}{s} \cdot V_{ip} [/tex]
And the fact that :
[tex] V_{out} = H(s) \cdot V_{ip} [/tex]
So that the [itex] S [/itex] in the numerator will disappear to form :
[tex] V_{out} = \frac{V_{ip} \times LR}{sLR + R_{fix}((LCR)s^2 + sL + R_{p})}[/tex]
This is one of the expressions I've derived, but I can't seem to hammer it into some useful form to take inverse laplace transforms of the second-shift nature, to form expressions which show a damped harmonic oscillations (which i know happens).
In a word...HELP!
If you can spot any obvious mistakes then please point them out, but I would be very grateful for anyone who can get me to a final solution for [itex] V(t) [/itex]
Cheers people
Brendan
I'm trying to derive an expression for the transient response (for a step input of magnitude V), for a non-ideal inductor modeled in the schematic I have drawn. This non-ideal inductor includes its inductance ( ), a parasitic parallel resistance and a parasitic capacitance.
The schematic is here: http://homepage.ntlworld.com/b.preece/RLC.JPG
So my initial thoughts are:
[tex]Z_{c}(S) = \frac{1}{cS}[/tex]
[tex]Z_{l}(S) = lS[/tex]
[tex]Z_{r}(S) = R[/tex]
Right... so the parallel combination of all of these three operational impedances will be defined as:
[tex]Z_{eff} = \left[\frac{1}{Z_{r}} + \frac{1}{Z_{c}} + \frac{1}{Z_{l}}\right]^{-1}[/tex]
So using those values quoted up above, we should have :
[tex]Z_{eff} = \left[\frac{1}{R} + Cs + \frac{1}{sL}\right]^{-1}[/tex]
Which if I'm not mistaked is evaluated to:
[tex] Z_{eff} = \frac{sLR}{(LCR)s^2 + sL + R_p} [/tex]
[itex] R_p [/itex] Any subscript 'p' indictating that its a parasitic value.
Ok... Now based upon the Voltage divider equation, we can say that the transfer function is as below:
[tex] H(s) = \frac{V_{out}}{V_{in}} = \frac{Z_{eff}}{Z_{eff} + Z_{fix}} [/tex]
Now we have:
[tex]{Z_{eff} + Z_{fix} = \frac{sLR + R_{fix}((LCR)s^2 + sL + R_{p})}{(LCR)s^2 + sL + R_p}[/tex]
And from there we have the Transfer function [itex]H(s)[/itex] to be:
[tex] H(s) = \frac{sLR}{sLR + R_{fix}((LCR)s^2 + sL + R_{p}}[/tex]
Knowing that the Laplace transform of the unit step function at t = 0, multiplied by a scaling value will be:
[tex] L(u(t)) = \frac{1}{s} \cdot V_{ip} [/tex]
And the fact that :
[tex] V_{out} = H(s) \cdot V_{ip} [/tex]
So that the [itex] S [/itex] in the numerator will disappear to form :
[tex] V_{out} = \frac{V_{ip} \times LR}{sLR + R_{fix}((LCR)s^2 + sL + R_{p})}[/tex]
This is one of the expressions I've derived, but I can't seem to hammer it into some useful form to take inverse laplace transforms of the second-shift nature, to form expressions which show a damped harmonic oscillations (which i know happens).
In a word...HELP!
If you can spot any obvious mistakes then please point them out, but I would be very grateful for anyone who can get me to a final solution for [itex] V(t) [/itex]
Cheers people
Brendan
Last edited by a moderator: