lecture 1 - filters and time response

7/30/2019 Lecture 1 - Filters and Time Response

http://slidepdf.com/reader/full/lecture-1-filters-and-time-response 1/13

MEM 640 Lecture 1: Filters and Time Response



Filters: Why Study Them?

RCVi oV

• Low Pass Filter is a classic first-order system• Easily assembled for experimentation• Differential equations are easy to derive: time domain• Frequency domain easy to derive• Show duality: time and frequency domain



Integrating Equation (3) yields:

11ln K t

RC V +−= Where 1K is a constant

Or

RC t o

K RC t eV eet V / / 1)( −− == Where oV is a constant (4)

RCt [sec]

Vo

Vo0.37

Plot of Equation (3) looks like:

Suppose RC t = then (3) yields: oo V eV RC V 37.0)(1

==−

At oo V eV RC V 05.0)3( 3 == −Engineers call the 95% value 3 time constants



Slightly Different Scenario

RCVi oV

Case Study: What is the voltage across the capacitor when the switch is closed?

The equation for this circuit is: R

V V dt

dV C I oio −≡= where iV is constant (5)

Then ( ) RC V

RC V

V V RC dt

dV oioi

o −=−= 1 (6)

The solution to the non-homogeneous differential equation (6) is:

RC t io AeV t V / )( −+= (7)

Given that 0)0( =oV then iV A −= and the solution to (7) becomes

)1()(/ RC t

io eV t V −

−=(8)



A plot of Equation (8) is

)1()( / RC t

ioeV t V −−=

t [sec]

Vi

Vo

Note:ii V eV RC V 63.0)1()( 1 =−= −

ii V eV RC V 99.0)1()5( 5 =−= −and

A first order system reaches 99% of steady-state in 5 time constants.Also note that the plot of Equation (8) looks like ramp (in early part). Thatis why a low-pass filter is also known as an integrator



Laplace Domain: Insights into Frequency Response

From (8) we derived the time response for a low pass filter as:

)1()( at io eV t V −−= where say

RC a

1= (9)

Taking Laplace transform of (9) yields

)()( assaV ii

oi

asV

sV

sV +=+

−= (10)

Now since iV Is a step input, recognize that

sV

asa

sV io +

=)(

Actual transferfunction

Step inputcontribution

Thus transfer function for a low pass filter is given as

asa

V V

i

o

+=(11)



as

a

V

V

i

o

+=

We will see later that the Bode Plot for (11) looks like

dB

frequency

-20 dB/decade slope

cutoff frequency

-3 dB

Recall that xdB log20= thus at -3 dB we have 707.010 20 / 3 == − x

Also recall from (8) that one time constant yields 63.3% of steady state

Thus the cutoff frequency is approximately the system’s time constant value

Also note that ( ) 499.0707.0 2 =

Power is proportional to voltage squared. Hence the -3 dB point represents

a 50% drop in power



Where are we going with this?

Problem: Show mathematically, the integrative properties of the LP filter

RCVi oV

Solution: The voltage across the resistor is oi V V − So:

R

V V

dt

dV C I oio −≡= (12)

Now suppose that io V V << then (12) can be re-expressed as

RV

dt dV

C ino ≈ (13)

Or,∫ ∫ +==

t

iin

o dt t V RC

dt RC V

t V constant)(1

)( (14)

The approximation (13) says the current is proportional to iV

. If iV is largeand is large, then we have a current source. R

Integration!



Hi Pass Filters Perform DerivativesProblem: Show that a hi pass filter perform differentiation

RCVi oV

Solution: The voltage across the cap is so we haveoi V V −

( ) R

V V V

dt

d C

dt

dV C I o

oi ≡−== (15)

Now if and chosen small enough so that R C

dt dV

dt dV oi >> (16)

Then R

V

dt

dV C oi ≈ (17)

Derivative!

Hencedt

dV RC V i

o = (18)



Grandmother Explanation



Based on Target Dynamics

• Shooter adds angle theta

• Adds phase

• Shooter angle LEADS target angle• Compensation

• Derivatives: rate of change

• Ignore high frequencies (hi-pass)

Differentiation LEAD compensation Hi-Pass Filter≈ ≈

lecture 1 - filters and time response

Documents