Digital filters are the most typical and popular DSP (or DTSP) systems. We treat and apply them in successive chapters. The objective of this section is to introduce briefly types of filters with associative difference equations. As for analog filters, digital filters act upon input signals to produce output signals which differ in amplitude, frequency and phase from those of the input signals. For the classification of frequency characteristic, digital filters, like analog counterparts, consists of 4 types: Lowpass, highpass, bandpass, and bandstop (or bandsuppress)( Figure ). The frequency aspect of filters will be discussed in next chapters when the Fourier analysis has been introduced. In this section, we will classify digital filters on basis of their impulse responses and their structures.

Nonrecursive filters and FIR filters In a nonrecursive filter the present output signal depends only on the input signal at all time. Mathematically, nonrecursive filters are described by the signal difference equation y(n)=∑k=−MMbkx(n−k) size 12{y \( n \) = Sum cSub { size 8{k= - M} } cSup { size 8{M} } {b rSub { size 8{k} } x \( n - k \) } } {} where bk size 12{b rSub { size 8{k} } } {} are filter coefficients which may be real or complex (usually assumed real). The limits are written as -M and M, but may be anything else, including −∞ size 12{ - infinity } {} and ∞ size 12{ infinity } {}. For causal filters, the lower limit is o. Structurally, a nonrecursive filter consists of three types of operators: Delay, multiplier and adder (summer). shows the direct implementation of a causal nonrecursive filter having 4 coefficients y ( n ) = ∑ k = 0 3 b k ( n − k ) size 12{y \( n \) = Sum cSub { size 8{k=0} } cSup { size 8{3} } {b rSub { size 8{k} } \( n - k \) } } {} In the figure z−1 size 12{z rSup { size 8{ - 1} } } {} is the unit delay (delay of one time index), z−2 size 12{z rSup { size 8{ - 2} } } {} is delay of two time indices, etc...

Structure (or implementation) of nonrecursive filter The impulse response h(n) size 12{h \( n \) } {} is obtained by replacing x(n−k) size 12{x \( n - k \) } {} by δ(n−k) size 12{δ \( n - k \) } {}: h(n)=∑k=−MMbkδ(n-k)=bn size 12{ size 10{h \( n \) = Sum cSub {k= - M} cSup {M} {b rSub {k} δ \( "n-k" \) } =b rSub {n} }} {} This result leads to two important conclusions: First the coefficient bn size 12{b rSub { size 8{n} } } {} is just the filter impulse response h(n) size 12{h \( n \) } {} at the same index n , second the nonrecursive filter defined by is also a Finite Inpulse Response (FIR) filter . Of course when one or both limits is/are ∞ size 12{ infinity } {}, we have no more a FIR filter but an IIR filter. However for most of the realistic cases the limits are finite, so nonrecursive filters and FIR filters are the same. When replacing bk size 12{b rSub { size 8{k} } } {}by h(k) size 12{h \( k \) } {} we have another form of nonrecursive (FIR) filters y(n)=∑k=−MMh(k)x(n−k) size 12{y \( n \) = Sum cSub { size 8{k= - M} } cSup { size 8{M} } {h \( k \) x \( n - k \) } } {} and for causal filters y(n)=∑k=0Mh(k)x(n−k) size 12{y \( n \) = Sum cSub { size 8{k=0} } cSup { size 8{M} } {h \( k \) x \( n - k \) } } {} Notice that nonrecursive filters implement directly the convolution summation. Consider a moving average filter of five terms y ( n ) = 1 5 [ x ( n − 2 ) + x ( n − 1 ) + x ( n ) + x ( n + 1 ) + x ( n + 2 ) ] size 12{y \( n \) = { {1} over {5} } \[ x \( n - 2 \) +x \( n - 1 \) +x \( n \) +x \( n+1 \) +x \( n+2 \) \] } {} which is a noncausal nonrecursive (or noncausal FIR) filter. When the time increases the averaging moves on, so is the name. The input signal is periodic with a period of 4 samples: x ( n ) = [ . . . 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3 . . . ] size 12{x \( n \) = \[ "." "." "." 1, matrix { } 2, matrix { } 3, matrix { } 2, matrix { } 1, matrix { } 2, matrix { } 3, matrix { } 2, matrix { } 1, matrix { } 2, matrix { } 3 "." "." "." \] } {} Plot the input and output signals. Solution The evaluation of the output signal proceeds as follows: y ( 0 ) = 0 . 2 [ x ( − 2 ) + x ( − 1 ) + x ( 0 ) + x ( 1 ) + x ( 2 ) ] = 0 . 2 [ 1 + 2 + 3 + 2 + 1 ] = 1 . 8 y ( 1 ) = 0 . 2 [ x ( − 1 ) + x ( 0 ) + x ( 1 ) + x ( 2 ) + x ( 3 ) ] = 0 . 2 [ 2 + 3 + 2 + 1 + 2 ] = 2 . 0 y ( 2 ) = 0 . 2 [ x ( 0 ) + x ( 1 ) + x ( 2 ) + x ( 3 ) + x ( 4 ) ] = 0 . 2 [ 3 + 2 + 4 + 2 + 3 ] = 2 . 2 y ( 3 ) = 0 . 2 [ x ( 1 ) + x ( 2 ) + x ( 3 ) + x ( 4 ) + x ( 5 ) ] = 0 . 2 [ 2 + 1 + 2 + 3 + 2 ] = 2 . 0 . . . alignl { stack { size 12{y \( 0 \) =0 "." 2 \[ x \( - 2 \) +x \( - 1 \) +x \( 0 \) +x \( 1 \) +x \( 2 \) \] =0 "." 2 \[ 1+2+3+2+1 \] =1 "." 8} {} # size 12{y \( 1 \) =0 "." 2 \[ x \( - 1 \) +x \( 0 \) +x \( 1 \) +x \( 2 \) +x \( 3 \) \] =0 "." 2 \[ 2+3+2+1+2 \] =2 "." 0} {} # size 12{y \( 2 \) =0 "." 2 \[ x \( 0 \) +x \( 1 \) +x \( 2 \) +x \( 3 \) +x \( 4 \) \] =0 "." 2 \[ 3+2+4+2+3 \] =2 "." 2} {} # size 12{y \( 3 \) =0 "." 2 \[ x \( 1 \) +x \( 2 \) +x \( 3 \) +x \( 4 \) +x \( 5 \) \] =0 "." 2 \[ 2+1+2+3+2 \] =2 "." 0} {} # size 12{ "." "." "." } {} } } {} The input and output and signals are shown in .

Example (a) x(n), (b) y(n) Notice that the moving average filter is a lowpass filter because it tends to reduce the variation in amplitude of the input signal. Of course, not all nonrecursive (or FIR) filters are lowpass. When the more samples are averaged the more the output signal is smooth, but also the more is the processing time. The moving average filtering bears a fundamental meaning of DSP (or DTSP), i.e. the computation on the samples is just the digital signal processing. Notice that the given filter is noncausal because the present value of y(n) size 12{y \( n \) } {} depends on the two future values x(n=1) size 12{x \( n=1 \) } {} and x(n=2) size 12{x \( n=2 \) } {} of input signal. This cause no problem if we process the stored data. In real-time processing (RTP) signals must be causal. The given filter can be made causal by making all samples, except the present, become the past : y ( n ) = 1 5 [ x ( n ) + x ( n − 1 ) + x ( n − 2 ) + x ( n − 3 ) + x ( n − 5 ) ] size 12{y \( n \) = { {1} over {5} } \[ x \( n \) +x \( n - 1 \) +x \( n - 2 \) +x \( n - 3 \) +x \( n - 5 \) \] } {} Now the output signal y(n) size 12{y \( n \) } {} is the average value around the central sample x(n−2) size 12{x \( n - 2 \) } {} insteal around x(n) size 12{x \( n \) } {}as previously. The signal x ( n ) = sin 2πn 60 + sin 2πn 10 60 ≤ n ≤ 320 size 12{x \( n \) ="sin" { {2πn} over {"60"} } +"sin" { {2πn} over {"10"} } matrix { {} # {} # {} } "60" <= n <= "320"} {} is applied to a nonrecursive (FIR) filter having impulse response h ( n ) = 0 . 1 0 ≤ n ≤ 9 = 0, otherwise alignl { stack { size 12{h \( n \) =0 "." 1 matrix { {} # {} } 0 <= n <= 9} {} # size 12{ matrix { {} # {} # {} } =0, matrix { {} # {} } ital "otherwise"} {} } } {} Plot the input signal x(n) size 12{x \( n \) } {}and the output signal y(n) size 12{y \( n \) } {}. This Example is adapted from Introductory Digital Signal Processing with Computer Applications, Revised Edition, Paul A. Lynn and Wolfgang Fuerst, John. Wiley & Sons, 1994. Solution The given input signal consists of a low frequency component having 60 samples in its 2π size 12{2π} {}period, and a high frequency component having 10 samples in its 2π size 12{2π} {}period. a depicts the signal. The given impulse response means the filter is a moving average filter of 10 samples of equal amplitude of 0.1. The equation of the filter is y ( n ) = ∑ k = 0 9 h ( k ) x ( n − k ) = h ( 0 ) x ( n ) + h ( 1 ) x ( n − 1 ) + . . . + h ( 9 ) x ( n − 9 ) = 0 . 1 [ x ( n ) + x ( n − 1 ) + . . . + x ( n − 9 ) ] alignl { stack { size 12{y \( n \) = Sum cSub { size 8{k=0} } cSup { size 8{9} } {h \( k \) x \( n - k \) } } {} # matrix { {} # {} # {} } =h \( 0 \) x \( n \) +h \( 1 \) x \( n - 1 \) + "." "." "." +h \( 9 \) x \( n - 9 \) {} # matrix { {} # {} # {} } =0 "." 1 \[ x \( n \) +x \( n - 1 \) + "." "." "." +x \( n - 9 \) \] {} } } {} Thus output samples from n=60 size 12{n="60"} {} towards future are y ( 60 ) = 0 . 1 [ x ( 60 ) + x ( 59 ) + . . . + x ( 51 ) ] y ( 61 ) = 0 . 1 [ x ( 61 ) + x ( 60 ) + . . . + x ( 52 ) ] y ( 62 ) = 0 . 1 [ x ( 62 ) + x ( 61 ) + . . . + x ( 53 ) ] alignl { stack { size 12{y \( "60" \) =0 "." 1 \[ x \( "60" \) +x \( "59" \) + "." "." "." +x \( "51" \) \] } {} # size 12{y \( "61" \) =0 "." 1 \[ x \( "61" \) +x \( "60" \) + "." "." "." +x \( "52" \) \] } {} # size 12{y \( "62" \) =0 "." 1 \[ x \( "62" \) +x \( "61" \) + "." "." "." +x \( "53" \) \] } {} } } {} b shows the output signal. The high frequency component is almost completely eliminaled whereas the low frequent component is left almost intact. Notice the transient response at the start: At the begining only x(n) size 12{x \( n \) } {}participates in the averaging, next x(n−1) size 12{x \( n - 1 \) } {} is added ... only after 10 averaging times we have enough 10 samples at each averaging and the output becomes stable (steady-state value).

Example (a) input signal, (b) output signal

Recursive filters and IIR filters At a recursive filter the output signal depends the input signal at all time and also on the previous output signal. The defining difference equation is y(n)=∑k=1Nakyn−k+∑k=−MMbkxn−k size 12{y \( n \) = Sum cSub { size 8{k=1} } cSup { size 8{N} } {a rSub { size 8{k} } y left (n - k right )+{}} Sum cSub { size 8{k= - M} } cSup { size 8{M} } {b rSub { size 8{k} } x left (n - k right )} } {} where ak size 12{a rSub { size 8{k} } } {}are coefficients for the recursive (feedback) part, and bk size 12{b rSub { size 8{k} } } {}are coefficients for the nonrecursive (forward) part. In theory, the limits M, N can be up to ∞ size 12{ infinity } {} but in reality they are finite. The limit N is the order of the filter. When all the coefficients ak size 12{a rSub { size 8{k} } } {} are zero we have a nonrecursive filter as before. Some authors write the difference equation in the form different from above, for example ∑ k = 0 N a k y ( n − k ) = ∑ k = − M M b k x ( n − k ) size 12{ Sum cSub { size 8{k=0} } 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 \( n - k \) } } } {} We stick to the former equation. We have taken the limits M, N as finite then the impulse response of the filter is also finite? Actually, the impulse response is, in general, infinite, so recursive filters are mainly of the IIR type. (see subsequent examples). It’s a very good idea to look at the structure (or implementation) of the filter. For this let’s take for example the difference equation. y ( n ) = 0 . 8y ( n − 1 ) − 0 . 5y ( n − 2 ) + 2x ( n − 1 ) + 4x ( n + 1 ) size 12{y \( n \) =0 "." 8y \( n - 1 \) - 0 "." 5y \( n - 2 \) +2x \( n - 1 \) +4x \( n+1 \) } {} is the most direct way (there are several direct ways), and the easiest to understand, to realize the filter.The problem of filter stucture will be discussed in a seperate chapter because there are various structures with different characterstics and for different applications.

The direct implementation of the IIR filter in example For a nonrecursive filter alone its coefficients bk size 12{b rSub { size 8{k} } } {} are the impulse response samples. But in a recursive filter the coefficiens bk size 12{b rSub { size 8{k} } } {}are no more so, and the impulse tesponse is found from the definition of impulse response, i.e., we replace x(n−k) size 12{x \( n - k \) } {}by unit impulse δ(n−k) size 12{δ \( n - k \) } {} and y(n−k) size 12{y \( n - k \) } {} by h(n−k) size 12{h \( n - k \) } {} to find the impulse response h(n) size 12{h \( n \) } {}(see section ). Consider a causal moving average filter (nonrecursive filter) of 10 indices y ( n ) = 0 . 1 [ x ( n ) + x ( n − 1 ) . . . x ( n − 9 ) ] size 12{y \( n \) =0 "." 1 \[ x \( n \) +x \( n - 1 \) "." "." "." x \( n - 9 \) \] } {} Find the equivalent recursive filter and its impulse response. Solution First let’s form the output at y(n−1) size 12{y \( n - 1 \) } {}: y ( n − 1 ) = y ( n − 1 ) + 0 . 1 [ x ( n − 1 ) + x ( n − 2 ) + . . . + x ( n − 10 ) ] size 12{y \( n - 1 \) =y \( n - 1 \) +0 "." 1 \[ x \( n - 1 \) +x \( n - 2 \) + "." "." "." +x \( n - "10" \) \] } {} Next let’s form the difference y ( n ) − y ( n − 1 ) = 0 . 1 [ x ( n − 10 ) ] size 12{y \( n \) - y \( n - 1 \) =0 "." 1 \[ x \( n - "10" \) \] } {} Thus y ( n ) = y ( n − 1 ) + 0 . 1 [ x ( n ) − x ( n − 10 ) ] size 12{y \( n \) =y \( n - 1 \) +0 "." 1 \[ x \( n \) - x \( n - "10" \) \] } {} which is the desired recursive filter. Its impulse response is given as h ( n ) = h ( n − 1 ) + 0 . 1 [ δ ( n ) − δ ( n − 10 ) ] size 12{h \( n \) =h \( n - 1 \) +0 "." 1 \[ δ \( n \) - δ \( n - "10" \) \] } {} Iterative computation will give h ( 0 ) = h ( 1 ) = h ( 2 ) = . . . = h ( 9 ) = 0 . 1 size 12{h \( 0 \) =h \( 1 \) =h \( 2 \) = "." "." "." =h \( 9 \) =0 "." 1} {} Above example shows that the recursive filter is more computational efficient than its nonrecursive counterpart since we need to compute much less terms. But recursive (IIR) filter may be unstable if the coefficients are not chosen properly. A causal recursive filter is represented by the difference equation y ( n ) = y ( n − 1 ) + 0 . 8y ( n − 2 ) + x ( n − 1 ) size 12{y \( n \) =y \( n - 1 \) +0 "." 8y \( n - 2 \) +x \( n - 1 \) } {} Find its impulse response. Solution The difference equation of the impulse response is h ( n ) = h ( n − 1 ) + 0 . 8h ( n − 2 ) + x ( n ) size 12{h \( n \) =h \( n - 1 \) +0 "." 8h \( n - 2 \) +x \( n \) } {} Because the filter is causal, we need to compute h(n) for n ≥ 0 only: h ( 0 ) = h ( − 1 ) + 0 . 8h ( − 2 ) + δ ( 0 ) = 1 h ( 1 ) = h ( 0 ) + 0 . 8h ( − 1 ) + δ ( 1 ) = 1 h ( 2 ) = h ( 1 ) + 0 . 8h ( 0 ) + δ ( 2 ) = 1 + 0 . 8 h ( 3 ) = h ( 2 ) + 0 . 8h ( 1 ) + δ ( 3 ) = 1 + 0 . 8 + 0 . 8 . . . h ( n ) = 1 + ( n − 1 ) 0 . 8 n > 0 alignl { stack { size 12{h \( 0 \) =h \( - 1 \) +0 "." 8h \( - 2 \) +δ \( 0 \) =1} {} # size 12{h \( 1 \) =h \( 0 \) +0 "." 8h \( - 1 \) +δ \( 1 \) =1} {} # size 12{h \( 2 \) =h \( 1 \) +0 "." 8h \( 0 \) +δ \( 2 \) =1+0 "." 8} {} # size 12{h \( 3 \) =h \( 2 \) +0 "." 8h \( 1 \) +δ \( 3 \) =1+0 "." 8+0 "." 8} {} # size 12{ "." "." "." } {} # size 12{h \( n \) =1+ \( n - 1 \) 0 "." 8 matrix { {} # {} # {} } n>0} {} } } {} Notice the impulse response is of infinite duration. Concerning the convolution, the computation of the output signal of a FIR filter is always possible, but because the IIR filter has impulse response of infinite duration, so how can we compute the output signal? (see section ).

Block processing Depending on applications and available facilities (software, hardware), one uses on-line or off-line processing. In on-line processing or sample processing , the present sample is processed together with previous samples which have been stored, the next sample is processed together with the samples before it except the oldest one. Sample processing is real-time processing (RTP) , it requires high speed, and, thus, is usually implemented by hardware (such as Digital Signal Processors). When the samples come continuously and our processing system is not fast enough, we have to store the coming data in the memory and process them later. This is off-line processing, or batch processing. Still, there is a problem, that is we cannot process a too long sequence of samples. The solution is to apply block processing. Because we process stored data, the processing system can be causal as well as noncausal. Furthermore, block processing does not require too high a speed, therefore pure software (programmes) is adequate, exept the data acquisition (DAQ) part. Let’s Lx size 12{L rSub { size 8{x} } } {}be the sample block length, then the sample sequence is x ( n ) = [ x ( 0 ) , x ( 1 ) , . . . , x ( L x − 1 ) ] size 12{x \( n \) = \[ x \( 0 \) ,x \( 1 \) , "." "." "." ,x \( L rSub { size 8{x} } - 1 \) \] } {} Usulally the vector notation x is used intead of the above sequence form (x is a line matrix). The sample time is Tx=LxT=Lxfs size 12{T rSub { size 8{x} } =L rSub { size 8{x} } T= { {L rSub { size 8{x} } } over {f rSub { size 8{s} } } } } {} where T s is sampling period (sampling interval), fs size 12{f rSub { size 8{s} } } {} sampling frequency (sampling rate). is an example.

A sample block of length L x Now consider a causal FIR filter having impulse response (coefficients) of length Lh size 12{L rSub { size 8{h} } } {} h ( n ) = [ h ( 0 ) , h ( 1 ) , . . . , h ( L h − 1 ) ] size 12{h \( n \) = \[ h \( 0 \) ,h \( 1 \) , "." "." "." ,h \( L rSub { size 8{k} } - 1 \) \] } {}