mm3fc mathematical modeling 3 lecture 5
DESCRIPTION
MM3FC Mathematical Modeling 3 LECTURE 5. Times Weeks 7,8 & 9. Lectures : Mon,Tues,Wed 10-11am, Rm.1439 Tutorials : Thurs, 10am, Rm. ULT. Clinics : Fri, 8am, Rm.4.503. Dr. Charles Unsworth, Department of Engineering Science, Rm. 4.611 Tel : 373-7599 ext. 8 2461 - PowerPoint PPT PresentationTRANSCRIPT
MM3FC Mathematical Modeling 3LECTURE 5
Times
Weeks 7,8 & 9.Lectures : Mon,Tues,Wed 10-11am,
Rm.1439Tutorials : Thurs, 10am, Rm. ULT.
Clinics : Fri, 8am, Rm.4.503Dr. Charles Unsworth,Department of Engineering Science, Rm. 4.611
Tel : 373-7599 ext. 82461Email : [email protected]
This LectureWhat are we going to cover &
Why ?• Frequency Response of Simple Systems.
(1st Order Difference System = ‘High Pass System’ )(2nd Order Difference System = ‘Low Pass System’ )(Cascaded Systems)(L-point Running Average Filter)
• The Dirichlet Function.(needed to understand the frequency response of the L-point Running Average Filter)
First Difference System• The first difference system is :
y[n] = x[n] – x[n-1]
• Has coefficients bk={1,-1} with frequency response :
• Thus the magnitude response is :
)wjsin(+)wcos(-1=e -1=)wH( ˆ-jw
))wcos(-2(1=)w(sin+)w2cos(-)w(cos+1 =
(sin(w))+))wcos(-(1=)}wIm{H(+)}wRe{H( = |)wH(| ∴
)wsin(= )}wIm{H( ),wcos(-1= )}wRe{H( Now,
22
2222
… (5.1)
))wcos(-)/1w(sin( tan⇒
)})w)}/Re{H(w(Im{H(tan=)wH(∠
1-
-1
))wcos(-(1=/2)w(2sin Thus, /2,w =Let x
cos2x)-(12
1=xsin Now,
2
2
• The phase response is :
/2)w2sin(=/2))w(2(2sin= |)wH(| ⇒ 2
From the magnitude plot• The system completely removes the DC component at w = 0.
• However, the high frequencies up towards are preserved.
Thus, this filter is known as a ‘high pass’ filter.
From the phase plot• We can see linear phase over the preserved frequencies.
For both plots we can see only the frequency range 0 < w < need to be plotted.
And Magnitude is an EVEN function.And Phase is an ODD function.
The Simple Low-Pass FIR Filter• Believe it or not ! We did this earlier. The difference equation :
y[n] = x[n] + 2x[n-1] +x[n-2]
• Gave the frequency response of Example 1, Lecture 4:
• The Magnitude plot shows the DC and low frequencies are preserved.• And the high frequencies are removed.
w-jw-2jw-j )).ewcos(2+2(=e+e2+1=)wH(
Frequency Response for Cascaded Systems
• When 2 LTI systems are in cascade then we ‘convolve’ the individual impulse responses of each system together.
• The frequency response of 2 LTI systems in cascade is simply the ‘product’ of the individual frequency responses.
x[n] = ejwn H1[w]ejwn
LTI 1H1[w]
LTI 2H2[w]
y1[n]= H1[w]H2[w]ejwn
LTI 2H2[w]
LTI 1H1[n]
x[n] = ejwn H2[w]ejwn y2[n]= H2[w]H1[w]ejwn
= H1[w]H2[w]ejwn
LTI EquivalentH[w]
x[n] = ejwn y[n] = H[w]ejwn
)w()Hw(H ⇔ [n]h*[n]h
tionMultiplica ⇔ n Convolutio
2121
• Thus,
Example 1 : Two LTI systems have coefficients ak={1,-2} and bk={0,1,1}. Determine their cascaded frequency response, impulse response, difference equation and the co-efficients of an equivalent filter.
H1(w) = 1 – 2e-jw and H2(w) = e-jw + e-2jw
H(w) = H1(w)H2(w) = (1 – 2e-jw)(e-jw + e-2jw) = e-jw + e-2jw – 2e-2jw - 2e-3jw
= e-jw - e-2jw - 2e-3jw
Thus the cascaded impulse response is : h[n] = [n-1] – [n-2] –2[n-3]
Thus, the cascaded difference equation is : y[n] = x[n-1] – x[n-2] –2x[n-3]
The equivalent filter has co-efficients : ck = {0,1,-1,-2}( Quite handy if you have 3 or more cascaded systems)
… (5.2)
• The LTI Running average FIR system is defined as :
• Thus, the frequency response can be written as :
• We can derive the magnitude and phase of the system by making use of the series expansion formula :
Frequency Response of an L-point Running Average Filter
∑1-L
0=k
k]-x[nL
1=y[n]
∑1-L
0=k
kwj-eL
1=)wH(
∑1-L
0=k
Lk
-1
-1=
α
αα
… (5.3)
• By letting = e-jw, we can expand the frequency response, such that :
• Now,
• Where DL(w) is a well known function known as the ‘Dirichlet function’, where (L) is the order of the L-point running average filter.
1)/2-(Lwj-
/2wj-/2wj/2wj-
L/2wj-L/2wjL/2wj-
wj-
Lw-j1-L
0=k
kwj-
e /2)wLsin(
L/2)wsin( =
)e-(ee
)e-(ee
L
1=
e-1
e-1
L
1=e
L
1=)wH( ∑
/2)wLsin(
L/2)wsin( =)w(D where,
e )w(D=)wH(
L
1)/2-(Lw-jL
((
(
(
))
)
)… (6.4)
A Closer Look at the Dirichlet Function
• Consider what the frequency response would be for an 11-point running averager.
• Thus, H(w) is a product of the real amplitude function D11(w) and a complex exponential function e-j5w.( Remember, e-j5w has magnitude = 1 and phase = -5w )
• ‘Amplitude’ rather than ‘Magnitude’ is used to describe D11(w) because D11(w) can be –ve.
• We obtain a plot of the magnitude |H(w)| by taking the absolute value of D11(w).
• We shall consider the amplitude representation first because it is simpler to examine the properties of the amplitude.
e /2)w11sin(
11/2)wsin( =e )w(D=e )w(D=)wH( wj5-wj5-
111)/2-(11wj-
11 )(
• The amplitude plot of the 11-point running averager is shown below :
Important Features to note :1) D11(w) is periodic with period 2.2) D11(w) has a maximum value = 1, at w = 0.3) D11(w) decays as (w) increases, with smallest nonzero amplitude
at w =
1) D11(w) has zeros at nonzero multiples of 2/11( In General, DL(w) has zeros at nonzero multiples of 2/L)
• For completeness, we know the phase of the 11-point running averager is linear with gradient of –5w.
The Magnitude response • for the 11-point running
averager is the absolute value of D11(w) :
|H(w)| = |D11(w)|
• D11(w) has zeros at nonzero multiples of 2/11.
• And null frequencies at these points
The phase response is :
• More complicated than the linear function we saw before.
• As we must include the algebraic sign in the phase function that the magnitude |H(w)| = |D11(w)| discards.
• A closeup of one period shows, the phase has a discontinuity at every nulled frequency and is linear inbetween each discontinuity.
• Moreover, in the amplitude we see that the discontinuities in the phase occur where the sign of the Dirichlet function changes.• At each sign change, where (w) is +ve we have a + phase jump.• At each sign change, where (w) is -ve we have a - phase jump.• Thus, we can construct the phase from gradient & phase jump knowledge.
Phase jump of + at eachsign change for +ve w
Phase jump of - at eachsign change for –ve w Gradient = (L-1)/2