Now we use the time convolution property (or convolution theorem) to map the output y(n)y(n) size 12{y \( n \) } {} in time domain to its transform Y(ω)Y(ω) size 12{Y \( ω \) } {} in the frequency domain (Fig.3.28) :

Or

The frequency response H(ω)H(ω) size 12{H \( ω \) } {}is usually a complex quantily, so we write

where

and

∣H(ω)∣∣H(ω)∣ size 12{ lline H \( ω \) rline } {} is the magnitude response anf Φ(ω)Φ(ω) size 12{Φ \( ω \) } {} is the phase response. If the impulse response h(n)h(n) size 12{h \( n \) } {} is real-valued then

The frequency response of a system exists if the system is BIBO stable, i.e. (Equation (2.15))

Example 3.7.1

Find the frequency response of a system whose input-output difference equation is

y ( n ) = A [ x ( n − 2 ) + x ( n − 1 ) + x ( n ) + x ( n + 1 ) + x ( n + 2 ) ] y ( n ) = A [ x ( n − 2 ) + x ( n − 1 ) + x ( n ) + x ( n + 1 ) + x ( n + 2 ) ] size 12{y \( n \) =A \[ x \( n - 2 \) +x \( n - 1 \) +x \( n \) +x \( n+1 \) +x \( n+2 \) \] } {}

where A is a constant.

Solution

First the impulse response h(n)h(n) size 12{h \( n \) } {} is just the output y(n)y(n) size 12{y \( n \) } {} when the input is x(n)=δ(n)x(n)=δ(n) size 12{x \( n \) =δ \( n \) } {}, thus

h ( n ) = A [ δ ( n − 2 ) + δ ( n − 1 ) + δ ( n ) + δ ( n + 1 ) + δ ( n + 2 ) ] h ( n ) = A [ δ ( n − 2 ) + δ ( n − 1 ) + δ ( n ) + δ ( n + 1 ) + δ ( n + 2 ) ] size 12{h \( n \) =A \[ δ \( n - 2 \) +δ \( n - 1 \) +δ \( n \) +δ \( n+1 \) +δ \( n+2 \) \] } {}

It turns out that this impulse response is the same as the signal in Example 3.4.1, hence the frequency response of the system is

H ( ω ) = ∑ n = − 2 2 h ( n ) e − jωn = A ( 1 + 2 cos ω + 2 cos 2ω ) H ( ω ) = ∑ n = − 2 2 h ( n ) e − jωn = A ( 1 + 2 cos ω + 2 cos 2ω ) size 12{H \( ω \) = Sum cSub { size 8{n= - 2} } cSup { size 8{2} } {h \( n \) e rSup { size 8{ - jωn} } } =A \( 1+2"cos"ω+2"cos"2ω \) } {}

The system is a low-pass filter.

Example 3.7.2

A system has impulse response.

h ( n ) = 0 . 8 n u ( n ) h ( n ) = 0 . 8 n u ( n ) size 12{h \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) } {}

Plot the frequency responses HR(ω),HI(ω),∣H(ω)∣HR(ω),HI(ω),∣H(ω)∣ size 12{H rSub { size 8{R} } \( ω \) ,H rSub { size 8{I} } \( ω \) , lline H \( ω \) rline } {} and Φ(ω)Φ(ω) size 12{Φ \( ω \) } {}.

Solution

This problem is the same as Example 3.4.2. The frequency response is

H ( ω ) = ∑ n = − ∞ ∞ h ( n ) e − jn ω = ∑ n = 0 ∞ ( 0,8 e − jω ) n = 1 1 − 0 . 8 e − jω H ( ω ) = ∑ n = − ∞ ∞ h ( n ) e − jn ω = ∑ n = 0 ∞ ( 0,8 e − jω ) n = 1 1 − 0 . 8 e − jω size 12{H \( ω \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \( n \) e rSup { size 8{ - ital "jn"ω} } = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { \( 0,8e rSup { size 8{ - jω} } \) rSup { size 8{n} } } } = { {1} over {1 - 0 "." 8e rSup { size 8{ - jω} } } } } {}

In order to compute the real and imaginary frequency responses we write

H ( ω ) = 1 − 0 . 8 e jω ( 1 − 0 . 8 e − jω ) ( 1 − 0 . 8 e jω ) = 1 − 0 . 8 cos ω − j0 . 8 sin ω 1 . 64 − 1 . 6 cos ω H ( ω ) = 1 − 0 . 8 e jω ( 1 − 0 . 8 e − jω ) ( 1 − 0 . 8 e jω ) = 1 − 0 . 8 cos ω − j0 . 8 sin ω 1 . 64 − 1 . 6 cos ω size 12{H \( ω \) = { {1 - 0 "." 8e rSup { size 8{jω} } } over { \( 1 - 0 "." 8e rSup { size 8{ - jω} } \) \( 1 - 0 "." 8e rSup { size 8{jω} } \) } } = { {1 - 0 "." 8"cos"ω - j0 "." 8"sin"ω} over {1 "." "64" - 1 "." 6"cos"ω} } } {}

From this,

H R ( ω ) = 1 − 0 . 8 cos ω 1 . 64 − 1 . 6 cos ω H R ( ω ) = 1 − 0 . 8 cos ω 1 . 64 − 1 . 6 cos ω size 12{H rSub { size 8{R} } \( ω \) = { {1 - 0 "." 8"cos"ω} over {1 "." "64" - 1 "." 6"cos"ω} } } {}

H I ( ω ) = 0 . 8 sin ω 1 . 64 − 1 . 6 cos ω H I ( ω ) = 0 . 8 sin ω 1 . 64 − 1 . 6 cos ω size 12{H rSub { size 8{I} } \( ω \) = { {0 "." 8"sin"ω} over {1 "." "64" - 1 "." 6"cos"ω} } } {}

For the magnitude response ∣H(ω)∣∣H(ω)∣ size 12{ lline H \( ω \) rline } {} we’d better not go from these two components, but rather from the original expression of H(ω)H(ω) size 12{H \( ω \) } {}:

H ( ω ) = 1 ( 1 − 0 . 8 cos ω ) + j0 . 8 sin ω H ( ω ) = 1 ( 1 − 0 . 8 cos ω ) + j0 . 8 sin ω size 12{H \( ω \) = { {1} over { \( 1 - 0 "." 8"cos"ω \) +j0 "." 8"sin"ω} } } {}

then

∣ H ( ω ) ∣ = 1 [ ( 1 − 0 . 8 cos ω ) 2 + ( 0 . 8 sin ω ) 2 ] 1 2 = 1 [ 1 . 64 − 1 . 60 cos ω ] 1 2 ∣ H ( ω ) ∣ = 1 [ ( 1 − 0 . 8 cos ω ) 2 + ( 0 . 8 sin ω ) 2 ] 1 2 = 1 [ 1 . 64 − 1 . 60 cos ω ] 1 2 size 12{ lline H \( ω \) rline = { {1} over { \[ \( 1 - 0 "." 8"cos"ω \) rSup { size 8{2} } + \( 0 "." 8"sin"ω \) rSup { size 8{2} } \] rSup { size 8{ { {1} over {2} } } } } } = { {1} over { \[ 1 "." "64" - 1 "." "60""cos"ω \] rSup { size 8{ { {1} over {2} } } } } } } {}

The phase response is

Φ ( ω ) = − tan − 1 0 . 8 sin ω 1 − 0 . 8 cos ω Φ ( ω ) = − tan − 1 0 . 8 sin ω 1 − 0 . 8 cos ω size 12{Φ \( ω \) = - "tan" rSup { size 8{ - 1} } { {0 "." 8"sin"ω} over {1 - 0 "." 8"cos"ω} } } {}

Fig.3.29 presents all the required spectra.

So far the linear scale has been used on the vertical axis (ordinates) to express the of frequency response magnitude. We know in electronics that the logarithmic scale, mainly the decibel (dB) scale, is often used for the horizontal axis (abscissa) to reduce large variations such as from 10 to 109109 size 12{"10" rSup { size 8{9} } } {}. But in DSP (DTSP) the logarithmic scale is used for ordinates to enlarge small variations in amplitudes to make the sidelobes more pronounced.

The magnitude in dBs H(ω)∣dBH(ω)∣dB size 12{H \( ω \) \lline rSub { size 8{ ital "dB"} } } {} is related to the magnitude ∣H(ω)∣∣H(ω)∣ size 12{ lline H \( ω \) rline } {} in linear scale as

See Fig.3.30. Remember when ∣H(ω)∣=1∣H(ω)∣=1 size 12{ lline H \( ω \) rline =1} {} then ∣H(ω)∣dB=0∣H(ω)∣dB=0 size 12{ lline H \( ω \) rline rSub { size 8{ ital "dB"} } =0} {}, when ∣H(ω)∣>1∣H(ω)∣>1 size 12{ lline H \( ω \) rline >1} {} the dBs are positive, when ∣H(ω)∣<1∣H(ω)∣<1 size 12{ lline H \( ω \) rline <1} {} the dBs are negative. Some examples are as follows.

Observing the logarith variation we see that when ∣H(ω)∣∣H(ω)∣ size 12{ lline H \( ω \) rline } {} is in the range 0 to 0.1, the dB varies extremely fast from −∞−∞ size 12{ - infinity } {} to – 20 dB. And when ∣H(ω)∣∣H(ω)∣ size 12{ lline H \( ω \) rline } {} has values from 0.5 upwards the dB slows down. Fig.3.30 materializes this observation. The small variation of ∣H(ω)∣∣H(ω)∣ size 12{ lline H \( ω \) rline } {}around 1 dissappears on ∣H(ω)∣dB∣H(ω)∣dB size 12{ lline H \( ω \) rline rSub { size 8{ ital "dB"} } } {}, whereas the unseen variation of ∣H(ω)∣∣H(ω)∣ size 12{ lline H \( ω \) rline } {} around 0 is greatly magnified on ∣H(ω)∣dB∣H(ω)∣dB size 12{ lline H \( ω \) rline rSub { size 8{ ital "dB"} } } {}.

Here, the idea is find a signal which preserves its time identity when going through a system. Let’s start with a discrete cosine

x ( n ) = cos ωn x ( n ) = cos ωn size 12{x \( n \) ="cos"ωn} {}

The corresponding signal out of a system represented by the inpulse response h(n) is

y ( n ) = ∑ k = − ∞ ∞ h ( k ) cos ω ( n − k ) = ∑ k = − ∞ ∞ h ( k ) cos ωk cos ωn + ∑ k = − ∞ ∞ h ( k ) sin ωk sin ωn y ( n ) = ∑ k = − ∞ ∞ h ( k ) cos ω ( n − k ) = ∑ k = − ∞ ∞ h ( k ) cos ωk cos ωn + ∑ k = − ∞ ∞ h ( k ) sin ωk sin ωn alignl { stack { size 12{y \( n \) = Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h \( k \) "cos"ω \( n - k \) } } {} # matrix { {} # {} } = left [ Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h \( k \) "cos"ωk} right ]"cos"ωn+ left [ Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h \( k \) "sin"ωk} right ]"sin"ωn {} } } {}

Both factors in brackets are independent of time as expected but there is an unwanted accompanied sine term. Now let’s test with a complex exponential

The output is

y ( n ) = ∑ k = − ∞ ∞ h ( k ) e jω n − k = ∑ k = − ∞ ∞ h ( k ) e − jωk e jωn y ( n ) = ∑ k = − ∞ ∞ h ( k ) e jω n − k = ∑ k = − ∞ ∞ h ( k ) e − jωk e jωn size 12{y \( n \) = Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h \( k \) e rSup { size 8{jω left (n - k right )} } } = left [ Sum cSub { size 8{k= - infinity } } cSup { size 8{ infinity } } {h \( k \) e rSup { size 8{ - jωk} } } right ]e rSup { size 8{jωn} } } {}

So the input exponential appears wholly at the output, its time variation does not change. The factor in brackets is just the frequency response H(ω)H(ω) size 12{H \( ω \) } {}, so

In the mathematical language H(ω)H(ω) size 12{H \( ω \) } {}is the eigen-value and ejωnejωn size 12{e rSup { size 8{jωn} } } {} the eigen-function. Actually, the phase of the input signal ejωnejωn size 12{e rSup { size 8{jωn} } } {} has been change. For this we write

y ( n ) = ∣ H ( ω ) ∣ e jΦ ω e jωn = ∣ H ( ω ) ∣ e j ωn + Φ ω y ( n ) = ∣ H ( ω ) ∣ e jΦ ω e jωn = ∣ H ( ω ) ∣ e j ωn + Φ ω size 12{y \( n \) = lline H \( ω \) rline e rSup { size 8{jΦ left (ω right )} } e rSup { size 8{jωn} } = lline H \( ω \) rline e rSup { size 8{j left (ωn+Φ left (ω right ) right )} } } {}

Example 3.7.3

A filter has impulse response

h ( n ) = 0 . 8 n u ( n ) h ( n ) = 0 . 8 n u ( n ) size 12{h \( n \) =0 "." 8 rSup { size 8{n} } u \( n \) } {}

Find the output for the input

Solution

First we find the filter frequency response

H ( ω ) = ∑ n = − ∞ ∞ h ( n ) e − jωn = ∑ n = 0 ∞ ( 0 . 8 e − jω ) n = 1 1 − 0 . 8 e − jω H ( ω ) = ∑ n = − ∞ ∞ h ( n ) e − jωn = ∑ n = 0 ∞ ( 0 . 8 e − jω ) n = 1 1 − 0 . 8 e − jω size 12{H \( ω \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {h \( n \) e rSup { size 8{ - jωn} } } = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { \( 0 "." 8e rSup { size 8{ - jω} } \) rSup { size 8{n} } } = { {1} over {1 - 0 "." 8e rSup { size 8{ - jω} } } } } {}

Notice the both signals in (a) and (b) have the same angular frequency of ω=π/2ω=π/2 size 12{ω= {π} slash {2} } {}. The frequency response at this frequency is

H π 2 = 1 1 + j0 . 8 = 1 1 − 0 . 8 e − jπ / 2 = 1 1 + j0 . 8 = 1 1 . 64 e − j 38 . 66 0 H π 2 = 1 1 + j0 . 8 = 1 1 − 0 . 8 e − jπ / 2 = 1 1 + j0 . 8 = 1 1 . 64 e − j 38 . 66 0 size 12{H left ( { {π} over {2} } right )= { {1} over {1+j0 "." 8} } = { {1} over {1 - 0 "." 8e rSup { size 8{ - jπ/2} } } } = { {1} over {1+j0 "." 8} } = { {1} over { sqrt {1 "." "64"} } } e rSup { size 8{ - j"38" "." "66" rSup { size 6{0} } } } } {}

(a) The output signal with respect to this input is

y ( n ) = H π 2 e jn π / 2 = 2 . 5 2 5 e − j 26 . 6 0 e jn π / 2 = 5 e j nπ / 2 − 26 , 6 0 − ∞ < n < ∞ y ( n ) = H π 2 e jn π / 2 = 2 . 5 2 5 e − j 26 . 6 0 e jn π / 2 = 5 e j nπ / 2 − 26 , 6 0 − ∞ < n < ∞ alignl { stack { size 12{y \( n \) =H left ( { {π} over {2} } right )e rSup { size 8{ ital "jn"π/2} } =2 "." 5 left ( { {2} over { sqrt {5} } } e rSup { size 8{ - j"26" "." 6 rSup { size 6{0} } } } right )e rSup { ital "jn"π/2} } {} # size 12{ matrix { matrix { {} # {} # {} # {} } {} # {} # {} } = sqrt {5} e rSup { size 8{j left (nπ/2 - "26",6 rSup { size 6{0} } right )} } matrix { {} # {} # {} } - infinity <n< infinity } {} } } {}

y ( n ) = H π 2 x ( n ) = 1 . 64 1 . 64 e j ( nπ / 2 − 38 . 66 0 ) = 1 . 64 e j ( nπ / 2 − 38 . 66 0 ) y ( n ) = H π 2 x ( n ) = 1 . 64 1 . 64 e j ( nπ / 2 − 38 . 66 0 ) = 1 . 64 e j ( nπ / 2 − 38 . 66 0 ) size 12{y \( n \) =H left ( { {π} over {2} } right )x \( n \) = { {1 "." "64"} over { sqrt {1 "." "64"} } } e rSup { size 8{j \( nπ/2 - "38" "." "66" rSup { size 6{0} } \) } } = sqrt {1 "." "64"} e rSup {j \( nπ/2} size 12{ - "38" "." "66" rSup {0} } size 12{ \) }} {}

(b) For the cosinusoidal input, first we write

2 cos n π 2 = e jn π 2 + e − jn π 2 2 cos n π 2 = e jn π 2 + e − jn π 2 size 12{2"cos"n { {π} over {2} } =e rSup { size 8{ ital "jn" { {π} over {2} } } } +e rSup { size 8{ - ital "jn" { {π} over {2} } } } } {}

Thus the output is

y ( n ) = H π 2 ejn π / 2 + H − π 2 e − jn π / 2 = 1 1 . 64 e − j 38 . 66 0 e jn π / 2 + 1 1 . 64 e j 38 . 60 0 e − jn π / 2 = 1 1 . 64 e j ( 38 . 60 0 − nπ / 2 ) + e − j ( 38 . 66 0 − nπ / 2 = 2 1 . 64 cos ( nπ / 2 − 38 . 66 0 ) y ( n ) = H π 2 ejn π / 2 + H − π 2 e − jn π / 2 = 1 1 . 64 e − j 38 . 66 0 e jn π / 2 + 1 1 . 64 e j 38 . 60 0 e − jn π / 2 = 1 1 . 64 e j ( 38 . 60 0 − nπ / 2 ) + e − j ( 38 . 66 0 − nπ / 2 = 2 1 . 64 cos ( nπ / 2 − 38 . 66 0 ) alignl { stack { size 12{y \( n \) =H left ( { {π} over {2} } right ) ital "ejn"π/2+H left ( - { {π} over {2} } right )e rSup { size 8{ - ital "jn"π/2} } } {} # matrix { {} # {} # {} } = { {1} over { sqrt {1 "." "64"} } } e rSup { size 8{ - j"38" "." "66" rSup { size 6{0} } } } e rSup { ital "jn"π/2} size 12{+ { {1} over { sqrt {1 "." "64"} } } e rSup {j"38" "." "60" rSup { size 6{0} } } } size 12{e rSup { - ital "jn"π/2} } {} # size 12{ matrix { {} # {} # {} } = { {1} over { sqrt {1 "." "64"} } } left [e rSup {j \( "38" "." "60" rSup { size 6{0} } - nπ/2 \) } size 12{+e rSup { - j \( "38" "." "66" rSup { size 6{0} } - nπ/2} } right ]} {} # size 12{ matrix { {} # {} # {} } = { {2} over { sqrt {1 "." "64"} } } "cos" \( nπ/2 - "38" "." "66" rSup {0} size 12{ \) }} {} } } {}

In various situations filters are connected in cascade or in parallel, and we would like to know the overall frequency response. By using the associativity and the distributivity of impulse responses, and the convolution theorem of DTFT we can obtain (Fig.3.34):

From the difference equation of general linear filter (Equation 2.21)

y ( n ) = ∑ k = 1 N a k y ( n − k ) + ∑ K = − M M b k x ( n − k ) y ( n ) = ∑ k = 1 N a k y ( n − k ) + ∑ K = − M M b k x ( n − k ) size 12{y \( n \) = Sum cSub { size 8{k=1} } cSup { size 8{N} } {a rSub { size 8{k} } y \( n - k \) } + Sum cSub { size 8{K= - M} } cSup { size 8{M} } {b rSub { size 8{k} } x \( n - k \) } } {}

For input ejωnejωn size 12{e rSup { size 8{jωn} } } {}the output is (Equation (3.70))

y ( n ) = H ( ω ) e jωn y ( n ) = H ( ω ) e jωn size 12{y \( n \) =H \( ω \) e rSup { size 8{jωn} } } {}

We replace this into the difference equation:

H ( ω ) e jn ω = ∑ k = 1 N a k H ( ω ) e jω n − k + ∑ k = − M M a k e jω n − k H ( ω ) e jn ω = ∑ k = 1 N a k H ( ω ) e jω n − k + ∑ k = − M M a k e jω n − k size 12{H \( ω \) e rSup { size 8{ ital "jn"ω} } = Sum cSub { size 8{k=1} } cSup { size 8{N} } {a rSub { size 8{k} } H \( ω \) e rSup { size 8{jω left (n - k right )} } +{}} Sum cSub { size 8{k= - M} } cSup { size 8{M} } {a rSub { size 8{k} } e rSup { size 8{jω left (n - k right )} } } } {}

Thus

This result means the when we know the coefficients of a filter we can write the expression of its frequency response immediately. Conversely, if we know the expression of the frequency response of a system we can write its difference equation.

The normal way to compute the frequency response in to express it as a rational function

Where ΦN(ω)ΦN(ω) size 12{Φ rSub { size 8{N} } \( ω \) } {} or ∠N(ω)∠N(ω) size 12{∠N \( ω \) } {} is the phase of N(ω)N(ω) size 12{N \( ω \) } {}, and ΦD(N)ΦD(N) size 12{Φ rSub { size 8{D} } \( N \) } {}or ∠D(ω)∠D(ω) size 12{∠D \( ω \) } {} is the phase of D(ω)D(ω) size 12{D \( ω \) } {}. The magnitude and phase responses are then given respectively by

Unless for simple filters (i.e. filters with a small number of coefficients) we must write a computer programme, or use a software such as Matlab, to compute the frequency response and especially the phase response.

Example 3.7.4

An IIR filter has these nonzero coefficients

a 0 = 0 . 04 , a 2 = − 0 . 05 , a 4 = 0 . 06 , a 6 = − 0 . 11 , a 8 = 0 . 32 , a 9 = − 0 . 5, a 10 = 0 . 32 , a 12 = − 0 . 11 , a 14 = 0 . 06 , a 16 = − 0 . 05 , a 18 = 0 . 04 a 0 = 0 . 04 , a 2 = − 0 . 05 , a 4 = 0 . 06 , a 6 = − 0 . 11 , a 8 = 0 . 32 , a 9 = − 0 . 5, a 10 = 0 . 32 , a 12 = − 0 . 11 , a 14 = 0 . 06 , a 16 = − 0 . 05 , a 18 = 0 . 04 alignl { stack { size 12{a rSub { size 8{0} } =0 "." "04", matrix { } a rSub { size 8{2} } = - 0 "." "05", matrix { } a rSub { size 8{4} } =0 "." "06", matrix { } a rSub { size 8{6} } = - 0 "." "11", matrix { } a rSub { size 8{8} } =0 "." "32", matrix { } a rSub { size 8{9} } = - 0 "." 5, matrix { } a rSub { size 8{"10"} } =0 "." "32",} {} # a rSub { size 8{"12"} } = - 0 "." "11", matrix { {} } a rSub { size 8{"14"} } =0 "." "06", matrix { {} } a rSub { size 8{"16"} } = - 0 "." "05", matrix { {} } a rSub { size 8{"18"} } =0 "." "04" {} } } {}

Write the difference equation of the filter and find the frequency response.

Solution

The difference equation the FIR filter is

y ( n ) = 0 . 04 x ( n ) − 0 . 05 x ( n − 2 ) + 0 . 06 x ( n − 4 ) − 0 . 11 x ( n − 6 ) + 0 . 32 x ( n − 8 ) − 0 . 5x ( n − 9 ) + 0 . 32 x ( n − 10 ) − 0 . 11 x ( n − 12 ) + 0 . 06 x ( n − 14 ) − 0 . 05 x ( n − 16 ) + 0 . 04 x ( n − 18 ) y ( n ) = 0 . 04 x ( n ) − 0 . 05 x ( n − 2 ) + 0 . 06 x ( n − 4 ) − 0 . 11 x ( n − 6 ) + 0 . 32 x ( n − 8 ) − 0 . 5x ( n − 9 ) + 0 . 32 x ( n − 10 ) − 0 . 11 x ( n − 12 ) + 0 . 06 x ( n − 14 ) − 0 . 05 x ( n − 16 ) + 0 . 04 x ( n − 18 ) alignl { stack { size 12{y \( n \) =0 "." "04"x \( n \) - 0 "." "05"x \( n - 2 \) +0 "." "06"x \( n - 4 \) - 0 "." "11"x \( n - 6 \) +0 "." "32"x \( n - 8 \) } {} # size 12{ matrix { {} # {} # {} } - 0 "." 5x \( n - 9 \) +0 "." "32"x \( n - "10" \) - 0 "." "11"x \( n - "12" \) +0 "." "06"x \( n - "14" \) } {} # size 12{ matrix { {} # {} # {} } - 0 "." "05"x \( n - "16" \) +0 "." "04"x \( n - "18" \) } {} } } {}

Because all the coefficients are real hence we can quickly find the expressions for the real and imaginary parts:

H R ( ω ) = 0 . 04 cos 0ω − 0 . 05 cos 2ω + 0 . 06 cos 4ω − 0 . 11 cos 6ω + 0 . 32 cos 8ω − 0 . 5 cos 9ω + 0 . 32 cos 10 ω − 0 . 11 cos 12 ω + 0 . 06 cos 14 ω − 0 . 05 cos 16 ω + 0 . 04 cos 18 ω H R ( ω ) = 0 . 04 cos 0ω − 0 . 05 cos 2ω + 0 . 06 cos 4ω − 0 . 11 cos 6ω + 0 . 32 cos 8ω − 0 . 5 cos 9ω + 0 . 32 cos 10 ω − 0 . 11 cos 12 ω + 0 . 06 cos 14 ω − 0 . 05 cos 16 ω + 0 . 04 cos 18 ω alignl { stack { size 12{H rSub { size 8{R} } \( ω \) =0 "." "04""cos"0ω - 0 "." "05""cos"2ω+0 "." "06""cos"4ω - 0 "." "11""cos"6ω+0 "." "32""cos"8ω} {} # matrix { {} # {} # {} } - 0 "." 5"cos"9ω+0 "." "32""cos""10"ω - 0 "." "11""cos""12"ω+0 "." "06""cos""14"ω {} # matrix { {} # {} # {} } - 0 "." "05""cos""16"ω+0 "." "04""cos""18"ω {} } } {}