Fundamentals of Digital Signal Processing

Transcription

1 University of Technology Department of Electrical Engineering Lect 1 Fundamentals of Digital Signal Processing Asst. Prof. Dr. Hadi T. Ziboon

2 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Digital Signal Processing and Computer Network Digital Signal Processing 1.1 Introduction Signal and systems representation, representation of signals in time, sampling and analog to signal conversion. 1.2 Time domain analysis Linear time invariant systems, impulse response and convolution sum. 1.3 Frequency Domain Analysis and Z-Transform Linear constant-coefficient difference equation, Fourier transform and frequency response, z-transform. 1.4 Discrete Fourier Transforms (DFT) and FFT Signal analysis and synthesis based on DFT, Fast Fourier Transform (FFT). 1.5 Filter Analysis Fundamental structures of digital filter. 1.6 Infinite Impulse Response (IIR) Digital filter 1.7 Finite Impulse Response (FIR) Digital Filter 1

3 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Computer Network 2.1 Introduction Internet architecture, OSI and TCP / IP reference models, network history and standardization, network topology, LAN, MAN and WAN. 2.2 Physical Layer Theoretical basis, various transmission medias, various well know networks, multiplexing, switching. 2.3 Data link layer Framing, error control (detection and correction), flow control. 2.4 Network layer Routing, network control, IP protocols, routing and control protocols. 2.5 Transport layer Reliable end end data transfer principles, flow control, end-end congestion control, UDP, TCP. 2.6 Application layer WWW, FTP, , DNS, Multimedia. 2.7 Network security Text Books [1]Introduction to digital signal processing with computer application, Paul Lynng [2] Digital Signal Processing, Li Tan [3] Data Communication and computer Network Behrous Feourozon,

4 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 1. Introduction 1.1 The scope of digital signal processing (DSP) 1.1 DSP techniques are now used to analyze and process signals and data arising in many areas of engineering, science, medicine, economics and the social sciences. 1.2 DSP is concerned with the numerical manipulation (treatment) of signals and data in sampled form. Using elementary operations as digital storage, delay, addition, subtraction and multiplication by constants, we can produce a wide variety of useful functions. For example to extract a wanted signal from unwanted noise, to assess the frequencies presented in a signal. 1.3 The general purpose computer can be used for illustrating DSP theory and application. However, if high speed real time signal processing is required, it may use special purpose digital hardware. Programmable microprocessors attached to a general purpose host computer. 1.4 Various terms are used to describe signals in the DSP environment. Discrete-time signal, we mean a signal which is defined only for a particular set of instants in time or sampling instants. 3

5 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 1.5 Discrete time signals may be divided into two categories:- a. Sampled data signals, which display a continuous range of amplitude values. b. Digital signals, in which the amplitude values are quantized in a series of finite steps. 1.6 Signals stored or processed in a computer are digital, because each sample value is represented by a finite length binary. 1.7 In practice, the term discrete time signal and digital are often used interchangeably. Fig (1.1) shows two discrete time signals x[n] y[n] x[n] DSP y[n] n=0 T nt n fig (1.1) Input and output signal of DSP processors x [n] is the sampled input to a digital signal processor. y [n] is the sampled output to a digital signal processor. T is the sample period. n denotes the sample number or index. 4

6 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 1.8 Fig (1.2) shows a typical DSP block diagram. Examples of input signals might be analog voltage temperature, pressure or light intensity. a. If the signal is not electrical, it is first converted to a proportional voltage variation by a suitable transducer. b. Analog filter, is used the first stage in the block diagram to limit the frequency range of the signal prior to sampling. c. The signal isnext sampled and converted into a binary code by an analog to digital converter (ADC). After digital signal processing it may be changed back into analog form by DAC. Analog filter is used the final stage to remove sharp transitions from the DAC output. d. After DSP the signal may be used to drive a computer display, or it Analog input may be transmitted in binary form to a remote terminal or location Some useful applications Fig (1.2) A typical DSP block diagram. 5 Analog output If the output of a typical DSP is given by y[n] for a given input signal x [n] Analog filter y[n]= F{x[n] + x[n-1] + x[n-2] x [n-n]} Constant or variable N ADC DSP DAC Analog filter y[n] = F { k=0 x[n k] } (1.1b) (1.1a)

7 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. y[n]= F{x[n] + x[n-1] + x[n-2] x [n-n]} N y[n] = F { k=0 x[n k] } (1.1) Equation (1.1):- a) is a recurrence formula, which is used over and over again to fined successive values of y. b) It is also called a difference equation. c) It is nonrecursive, because each output is computed simply form input values. d) It is used to define filters if each term multiplied by a constant or variable number [F]. 2.2 The above algorithm described by eq(1.1) is not very efficient because of each output value is exactly the same as the one before it, except that the most recent input sample is included. Eq (1.1) can be smoothed the output signal y[n] by y[n+1] = y[n] + F {x[n+1] x[n-n]}.. (1.2) Since eq (1.2) is a recurrence formula which applied for any value of n, we may subtract (1) from each term in square brackets, giving y[n] = y[n-1]+f{x[n] x[n-n-1]}.. (1.3) Eq (1.3):- a) Confirms that we can estimate each output sample by updating the previous output y [n-1]. b) The equation defines a recursive version of the filter which is more efficient, required fewer addition and subtraction. 6

8 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng Sampling and Analog to Digital Conversion Suppose that an analog signal is to be represented by a set of equally spaced samples. Fig (1.3.a) shows a signal sampled at a rate which is clearly too low to pick out the more rapid fluctuations. Fig (1.3.b), sampling appears unnecessarily fast. A great many samples values would have to be stored, processes or transmitted. x[n] x[n] n n Fig (1.3.a)Fig (1.3.b) Thus must choose suitable or intermediate sample rate which is adequate, without excessive. 3.2 Shannon ' S sampling theorem, which may be stated as follows:- An analog signal containing components up to some maximum frequency f1 Hzmay be completely represented by regularly spaced samples, provided the sampling rate is at least 2f1 samples per second. Thus a speech waveform having f1=3khz should be sampled at least 6000 Hz. If we use the minimum rate specified by the theorem, then the sampling interval T is given by 7 T = 1 2f 1... (1.4)

9 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng Alternatively, if we have a digital system with sampling interval T, the maximum analog frequency is called the Nyquist frequency and is given by f 1 = 1 2T Hz or W 1 = 2πf 1 = π T radians/s (1.5) 3.4 Fig (1.4a) represents the frequency distribution or spectrum of an analog speech signal. There are no components above maximum frequency. / H(f) / (a) f 1 f 1 = 3KHz 0 3f(KHz) / H(f) / Reconstructed filter (b) f s = 2f 1 = 6KHz aliasing / H(f) / 3 6 a aliasing (c) f s < 2f 1 = 5KHz f Fig (1.4) The effects of sampling on a signal spectrum. From fig (1.4) can be concluded:- a) The spectrum as an even function, extending to negative frequencies. This widely used representation results from expressing each frequency component as the sum of two exponentials. Thus a component Acoswt may be written as Acos(wt) = A 2 ejwt + A 2 e jwt. (1.6) b) Reconstructing filter is used to recover the original signal. The filter type is low pass filter. 8

10 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng In most electronic DSP applications, sampling is performed by an analog to digital converter (ADC), which also transforms the stream of samples into a binary code. The binary code N bit long allows2 N separate numbers. Thus if N=8, we may encode2 8 = 256, discrete values Analog signals normally take on a continuous range of amplitudes, so when they are sampled and binary coded small amplitude errors are bound to be introduced as shown in fig (1.5) with 3 bite code. Sample value x[n] Binary code Fig (1.5) Converting an analog 5 signal into a binary code (3bite) n Fig (1.5) shows the following note:- a) With 3 bit code, the separate sample values are 8. The total amplitude range is divided into 8 quantization levels or slots. b) The maximum error introduced into each sample value by this process is± half quantization level. With 3 bite code the error ± 1 available amplitude range. This error is called quantization noise. 16 of the total c) Analog to digital conversion always degrades a signal to some value. 9

11 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng After a signal has been processed digitally, it may be converted back to an analog voltage using a digital to analog converted (DAC). Fig (1.6) shows digital to analog conversion:- a) Each signal sample value, corresponding to the binary code delivered to the DAC input, is held drawing the following sampling interval. b) The resulting staircase waveform is suitable for many practical applications (including the computer plots). c) If a true analog output is required, a further smoothing filter must be employed in order to obtain fully reconstructed signal. Binary code Sample value n 7 5 DAC output 2 Time Analog signal Time Fig (1.6) Digital to analog conversion. 10

12 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng Basic type of digital signal Many DSP algorithms are linear. The response of a linear algorithm, or processor, to a number of signals applied simultaneously equals the summation of its responses to each signal applied separately. Thus if we can define the responses to each signal applied separately. Thus we can define the response of a linear processor to basic signals, we can predicate its response of a linear processor to basic signals; we can predict its response to more complication environment (summation of a number of simpler basic signals) Unit step function The unit step function u[n] is defined as u(n) u[n]=0 n<0 u[n]=1 n 0 ] (1.7) Fig (1.7) shows unit step 1.4.2Unite impulse function The unit impulse function δ[n] is shown in fig (1.8) and defined as δ[n] δ[n]=0 n 0 δ[n]=1 n=0 ] (1.8) fig (1.8) unit impulse function n 1 The relationship between unite step and unite impulse is given by u[n] = 11 n δ(m) m=. (1.9)

13 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Conversely, we can easily generate δ[n] from u[n]. The recurrence formula: Eq(1.10) holds good for all integer values of n Unit ramp function δ[n] = u[n] u[n-1] (1.10) The unit ramp function is shown in fig (1.9) and defined as r[n] = n u[n]. (1.11) since u[n] is zero for n<0 r[n] so also ramp function is zero for n<0 fig (1.9) The unit ramp function. Ex 1.1Find expression for the following signal.x[n] n a) This is a unit step function which has been scaled (Weighted) by of -2, it starts at n= - 4, rather than n=0, and it is time reversed. -2 b) Scaling by -2gives the function -2u[n] c) Time shifting so that it starts at n=-4 gives the function -2u[n=-4] d) Reversal gives the function -2u[-n-4].Hence the required function is x[n] = 2u[ n 4] since u[n] = 1 n > 0 = 2u[ (n + 4)] = 0 n < 0 in this case reversal 12 n

14 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex 1.2 Find expression for the following signal x[n] Solution a) This rectangular pulse may be considered 1 as the summation or superposition, of a unit step at n= -3 and an equal but opposite unite step at n= -5 n b) x[n] = u[n+3] u[n-5] Ex 1.3 Find an expression for the following signal Solution a) The signal is weighted unit impulse of value 8, time shifted to occurat n=6 x[n] 8 b) x [n] = 8 δ[n-6]n 6 Ex 1.4 Find expression for the following signal Solution a) This signal may be considered as x[n] the superposition of two ramp 6 8 Function: One starts at n=-6 and has slop of 2 n Its upward trend may be stopped by adding another ramp 2. The another ramp starting at n=-2 with slope of -2 x[n]= 2r[n+6]-2r[n+2] 13

15 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng Exponential, sine and cosines signal a) Exponential signals are used in analysis and often occur in the natural world and in technology. In the DSP context, sampled exponentials are used as a powerful set of frequency domain techniques to analyze and process signal and to design the system. The exponential signal is defined by x[n] = A e βn. (1.12) Where A and β are constants. b) Fig (1.10) shows the forms of signal for different values of β. x[n] x[n] x[n] β<0 β=0 β>0 n n n fig (1.10) Basic digital signal real exponentials. c) Such real exponentials theoretically continue forever in both directions. In practice, we work with defined signals in time such as x[n] = Aeβn n 0. (1.13) 0 n < 0 we can make use of the fact that the unit step u[n] is zero for n<0 and unity for n 0, if multiply or modulate an exponential by u[n], then x[n] x[n] = A e βn u[n]. (1.14) d) The successive sample values from a simple geometric progression each value equals that of its neighbor, multiplied by constant βn. x[n] = A e βn = AB n where B = e β Hence x[n + 1] = AB n+1 = [AB n ]B = Bx[n]. (1.15) 14

16 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. e) If we next allow the constant β in eq (1.12) to be purely imaginary. Let us considerβ=jω.. (1.16) Where j = 1, Ω= is a real constant. x 1 = A e βn = Ae jωn = A cos (Ωn) +ja sin(ωn).. (1.17) Eq (1.17) contains a cosin real part and sin imaginary part. However we deal mainly with real signals, if we put β= -jω, another signal is defined x 2 [n] = A e jω = A cos(ωn) jasin(ωn) f) By adding x 1, x 2 x 1 [n] + x 2 [n] = A cos(ωn) + ja sin(ωn) + A cos(ωn) ja sin(ωn) x 1 [n] + x 2 [n] = 2A cos(ωn). (1.18) A cos(ωn) = 1 2 x 1[n] x 2[n] = A 2 ej(ωn) + A 2 e j(ωn).. (1.19) This result shows that a sampled cosine signal can be made up from a pair of sampled imaginary exponentials g) By subtracting x 1 and x 2 x 1 x 2 = 2j sin(nω) A sin(nω) = A 2j ej(nω) A 2j e j(nω) (1.20) Therefore, a real sampled, sine signal may also be made us from a pair of imaginary exponentials. 15

17 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. What is the difference between sampled sine and cosines and their counterpart? 1) Analog sine and cosine signals are oscillatory and periodic sample sines and cosines, however, are not necessarily periodic. Although their sample values lie on a periodic envelope, the numerical values may not form a repetitive sequence. Fig (1.11a) shows a discrete time sinusoid and fig (1.11b) shows a discrete time cosinusoid both are periodic but fig (1.11c) shows a discrete time samples lies a long a sinusoidal envelope, does not have repeating numerical values. Exact repetition will only occur if the sampling interval bears some simpler relationship to the repetition time or period of the analog signal. This means that x[n] = A e jnω = A e j[n+n]ω = A e j(nω) e j(nω) (1.21) But e jnω =1 if NΩ is a multiple of 2π NΩ = 2πm or Ω 2π = m N (1.22) Hence x[n] is only periodic if Ω is a rational number (the ratio of two 2π integer). Otherwise its sample values do not repeat. 2) A second difference between analog and digital sinusoidal signals concerns the question of time and frequency scales. If x[n] = A sine [nω]. (1.23) n is assumed to be a dimensionless integer thus Ω in radians But one completes period corresponding to nω = 2π or n = 2π / Ω (1.24) thus 2π / Ω is sample/period regardless of the time or frequency. 16

18 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. x[n] x[n] n n (a) Sine x[n] (b) Cosine n (c) Discrete time sample lies along a sinusoidal envelope Fig (1.11) Basic digital signals sines and cosines. - If the sampling instants are given by t = n T, n= -2,-1,0,1,2, (where n= index represents time). (1. 24a) and the sampling frequency is f s = 1 T. (1.24b) The digital signal is given by x[n] = sin (nω) = sin (nwt). (1.25) since nt represents time in seconds, w must be an angular frequency in rad/sec. If f is corresponding frequency, then w = 2πf and eq (1.25) may be written as n 2πf x[n] = sin( ).. (1.26) f s Thus the sample series may be generated if both f and f s are know. 17

19 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex (1.5) Sketch carefully, and defined the values for each samples of the following signals:- a) x[n] = exp (0.2n) b) x[n] = cos( πn 4 ) x[n](1.22) c)x[n] = exp( n 15 ) sin(πn 6 ) 0.22 (a) n d) x[n] = exp n cos(n)u[n] (b) 5 n n n (c) (d) 1.5. Digital Processors Processor means any system which carriers out a DSP function. It may be a general purpose computer, microprocessor, special hardware or a combination of these. In general the processor is a computer plus software No particular distinction between the terms processor and system. 18

20 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng Linear Time Invariant (LTI) System Properties A linear system or processor may be defined as one which obeys the principle of superposition. If an input consisting of the sum of an number of signals is applied to a linear system, then the output is the sum or superposition of the system's response to each signal considered separately. Let as consider an input x 1 [n]applied to a digital processor produced y 1 [n]and that input x 2 [n] produced output y 2 [n]. Then the process are is linear if its response to {x 1 [n] + x 2 [n]} is {y 1 [n] + y 2 [n]} Furthermore, if an input a 1 x[n]produced output ay[n] where a is constant coefficient or multiplier or a weighting factor. In general the weighted sum of inputs a 1 x 1 [n] + bx 2 [n] + cx 3 [n] + must produce the weighted sum of outputs a 1 y 1 [n] + by 2 [n] + cy 3 [n] + Let us consider a system which squares each sample value applied to its input. Thus for two different inputs applied separately, we have y 1 = (x 1 [n]) 2 and y 2 = (x 2 [n]) 2. (1.27) When the same two inputs are summed, and applied simultaneously, the output is y 3 = {x 1 [n] + x 2 [n]} 2 = {x 1 [n]} 2 + {x 2 [n]} 2 + 2x 1 [n]x 2 [n] (1.28) 19

21 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Eq(1.28) is not the sum of its responses to x 1 [n] and x 2 [n] applied separately Frequency Preservation: - It means that if we apply an input signal containing certain frequencies to a linear system, the output can contain only the same frequencies and no others. The property depends upon the fact a sampled sinusoid, applied to any linear processor; produce a similar form of output. This property does not hold for the squaring system mentioned above. Suppose a signal sin(nω) is y[n] = {x 1 [n]} 2 = sin 2 [nω]. (1.29) = 1 2 {1 + cos[2nω]} = cos [2nΩ] 2 Thus there is no component in output at frequency = Ω Time Invariance:-A time-invariant system is one whose properties do not vary with time. The only effect of a time shift in an input signal to the system is a corresponding time shift in its output. The majority of technological systems and processes are of this type. 6.5 Association and Commutation Association property means that we may analyze a complicated LTI system by breaking it down into a number of simpler subsystems. Also, we can synthesize an overall system perhaps a very complicated one by designing a number of independent subsystems. 20

22 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. The commutative property of LTI systems means that if the subsystems are arranged in series or cascade, then they may be arranged in any order without affecting overall performance. This property advantages to reduce the complexity Other System Properties [Causality, Stability Invertibility and Memory] Causality: In a causal system, the output signal depends only on present and/or previous values of the inputs. Most practical signal processors are casual. However, if we record a signal on data and subsequently process by computer, the software need not be causal. Ex. y[n]=x[n]+x[n-1] causal y[n]=x[n]+x[n+1] noncausal 7.2 Stability: A stable system is one which produces a finite or bounded output in response to a bounded input Invertibility: If a digital processor with input x[n] gives an output y[n], then its inverse would produce x[n], if fed with y[n]. The most practical system is invertible. An exception in the squaring device mentioned earliest for which y[n] = {x[n]} Memory: A processor possesses memory if its present output y[n] depends upon one or more previous input values x[n-1], x[n-2],.. 21

23 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex (1.6) x[n] and y[n] are the input and output signals of a DSP system. Determine which of the following properties and possessed by systems defined by the recurrence formula (a) and (d) below:- Linearity Invertibility Stability Causality Time Invariance Memory a) y[n]= 3x[n] 4x[n-1] b) y[n] = 2y[n-1] + x[n+2] c) y[n] = n x[n] d) y[n] = cos[x[n]] Solution a) The output is a weighted sum of present and previous inputs - It is bounded if the input is bounded. - Thus the system has all six properties mentioned above. b) The present output depends on a future input, so the system is not causal. - If the input signal ceases, the output goes on rising without limit, since each output value twice the previous one. Therefore the system is unstable. - However a possess the other properties mentioned above namely: linearity, time-invariance, invertibility and memory. c) The output depends on the present input only, so the system has no memory. - Since it also depends on the independent variable, the system is time-variant - But the system have the properties of linearity, causality, stability and invertibility. 22

24 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. d) y[n] = cos[x[n]] A cosine function is periodic; so many different values of x[n] would produce the same value of y[n]. Hence the system is not invertible. - It is not a linear because if we double x[n], we do not double y[n]. - Since y[n] depends only on x[n], the system has no memory. - However, it is time invariant, causal and stable Classification of Signals Deterministic and Random Signal A signal can be classified as deterministic, meaning that there is no uncertainty with respect to its value at any time or as random that there is some degree uncertainty before the signal actually occurs Periodic and Nonperiodic Signal A signal is periodic in time if there exist a constant T 0 > o such that x(t) = x(t + T 0 ) where T 0 defines the duration of one period Analog and Discrete signals Analog signal x(t) is a continuous function of time, that is x(t) is defined for all t. By comparison a discrete signal x[nt] is one that exists only at discrete times, it is characterized by a sequence for each [nt]. 23

25 Lect. 1, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng Analog and Digital Signals a) Analog signal amplitude can take on an infinite number of values. b) Digital signals can take only a finite number of values. c) The term continuous time and discrete time define the nature of the signal along the time (horizontal axis). The terms digital and analog define the nature of the signal amplitude (vertical axis). Fig (1.11) shows examples of various types of signals. g(t) g(t) Analog continuous time time time Digital continuous time time time Analog discrete time Digital discrete time Energy and Power Signal a) A signal is defined an energy signal if it has nonzero but finite energy (0 < E x < ) for all time where E x = lim T 24 T/2 T/2 x 2 (x) dt (1.30) b) A signal is defined as a power signal if it has nonzero but finite power (0 < P x < ) for all time where P x = lim T 1 T/2 T T/2 x2 (x) dt (1.31) In general:- a) Periodic and random signal are classified as power signals. b) Deterministic and nonperiodic signal classified as energy signals.

26 University of Technology Department of Electrical Engineering Lect2 Time Domain Analysis Asst. Prof. Dr. Hadi T. Ziboon

27 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 2.Time Domain Analysis 2.1 Introduction The basic techniques for describing digital signals and processors in the time domain are:- a) Convolution:- allows us to find the output signal from any LTI processor in response to any input. b) Impulse response:- this is the response of the processor to the unit impulse response δ(n). The time domain methods are used for analysis and are not a great help in designing new processor. The main reason is that the design specifications are based on performance in the frequency domain. The convolution takes place in the time domain, without any need to consider the frequency components of the input signal and processor. 2.2 Describing Digital Signals with Impulse Function The digital signal x[n] is shown in fig(2.1) and it is clear that x[n] may be considered as the superposition or summation of the more basic impulse signals shown in parts x[1]δ[n-1] (e) fig(2.1) x[n] (a) n x[2]δ[n-2] x[-2]δ[n+2] (f) n n Each of these is a unit impulse shiftedx[-1]δ[n+1] x[-1]δ[n+1] by 'n' samples which has been weighted by the value of x[n] [ i.e the value of x[0]δ[n] x[n] for each n instant] (b) (c) (d) n n n n 1

28 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Thus each of these signal shown in fig(a-f) is a unit impulse which has been weighted by the appropriate value of x[n] [ i.e the value of x[n] is shown in part a], and shifted by a number of sampling intervals. The complete signal x[n] can be defined as x[n] = x[-2]δ(n+2)+ x[-1]δ(n+1)+ x[0]δ(0)+x[1]δ(n-1) +x[2]δ(n-2).. (2.1) or x[n] = x[k]δ[n k] k= (2.2) Where the integer K takes all value between ±, x[n] is a completely general digital signal. 2.3 Describing Digital LTI Processors The Impulse Response The aim of impulse response is to examine the nature of an LTI processor's response to an individual impulse. The impulse function is time limited signal δ[n]. δ[n] is finite at n=0, but zero elsewhere. If we deliver δ[n] to the input of digital processor, (the excitation) is confined to the instant n=0. Any output signal observed after n=0 must be characteristics of the processor itself. The impulse response is given by the symbol h[n]. If the input to a linear processor is δ[n], the output must be h[n] as shown in fig (2.2). From impulse response can be determined the system properties such as causality,stability and memory. 2

29 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. δ[n] Digital LTI processor h(n) fig (2.2) the impulse response of an LTI processor Fig (2.3) shows four examples of impulse response:- a) Shows a memory less system b) Shows a noncausal, since h(n) starts before n=0. c) Shows a unstable, because h[n] grows without limit as n increases. d) Shows a causal, stable, system with memory. (a) h[n] (b) h[n] memoryless noncausal (c)h[n] (d) h[n] Unstable causal, stable, with memory n n fig(2.3) various forms of impulse response Digital processor is often described by a recurrence formula or difference equation relating its input and output signals. It is easy to find the impulse response by this equation. Ex 2.1 Find the first three sample values of the impulse response h[n]for the system defined by the following difference equation y[n]= 1.5 y[n-1] y[n-2] + x[n]. (2.2) Assuming the system is a causal. 3

30 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Solution if the system causal, then h[n] h[n]=0 for n 0 and x[n] = δ[n] But δ(n)= 1 for n=0 and δ(n)=0 for elsewhere Substitute x[n]=δ[n] in eq(2.2) y[n]=1.5y[n-1]-.85y[n-2]+x[n]] h[n] = 1.5h[n 1] 0.85 h[n 2] + δ[n]. (2.3) It is simple to find the value of each term of h[n] for a given value of n If n=0, substitute n=0 in eq(2.3) h[0] = 1.5h[ 1] 0.85 h[ 2] + δ[0] = = 1 Since h[n]=0 for n 0 If n=1, substitute n=1 in eq(2.3) h[1] = 1.5h[0] 0.85 h[ 1] + δ[1] = = 1.5 If n=2, h[2] = 1.5h[1] 0.85 h[0] + δ[2] = = 1.4 and so on for other value of n Note: that is, in this example, the input impulse only contributes at n=0, but the value of h[n] depend entirely on previous values, being generated recursively. Program can be used to evaluate h[n] and their program in appendix A1 in the reference by Lynn. 4

31 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. EX 2.2 Find the first three sample values of the impulse response h[n] for the system shown in figure x(n) + y(n) Solution: T From the above figure, the Difference Equation can be written as y[n]= -0.9 y[n-1] +x[n] The impulse response is therefore given by:- h[n]= -0.9 h[n-1] + δ[n] The system is causal, so that h[n]=0 for n 0 and zero otherwise h[0] = 0.9 h[ 1] + δ[0] = =1 h[1] = 0.9 h[0] + δ[1] h[1] = = 0.9 h[2] = 0.9 h[1] + δ[2] h[2] = = 0.81 h[3] = 0.9 h[2] + δ[3] h[3] = = Thus each sample value is -0.9 times the previous one, and h[n] therefor follows decaying real, exponential envelope, with alternate sample. 5

32 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. EXP 2.3 find the first four sample values of the impulse response h[n] for the system defined by the following equation Solution:- y[n] = x[n] + x[n 1] + x[n 2] + This is a nonrecursive system, since no feedback in the equation The impulse response is therefor given by the system is causal h[n] = δ[n] + δ[n 1] + δ[n 2] + h[0] = δ[0] + δ[ 1] + δ[ 2] + h[0] = = 1 h[1] = δ[1] + δ[0] + δ[ 1] + = = 1 Similarly h[2] =h[3] =1 and so on. The impulse response therefor equals the unit step. EXP 2.4Repeat Ex. 2.3 if the equation y[n] = y[n 1] + x[n] Solution:-this is a recursive system, since there is a feedback in the equation The impulse response is given by Using the same procedure to find h[n] = h[n 1] + δ[n] h[0] = h[ 1] + δ[0] = = 1 h[1] = h[0] + δ[1] = = 1 h[2] = h[1] + δ[2] = = 1 h[3] = h[2] + δ[3] = = 1 6

33 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng The Step Response The unit step function u(n) is the running sum of the unit impulse δ[n],so the step response of a LTI processor is the running sum of its impulse response. If the step response is given by s[n], we have: or h[n] is the first order difference of s[n]: s[n] = n m= h[m]. (2.4) h[n] = s[n] s[n 1]. (2.5) A step response gives essentially the same information as an impulse response, but in slightly different form. Step response are useful for several reason:- 1- The step functions occur quite often in practice. 2- The response of a system to a sudden disturbance is evaluated by step response. 3- The process of convolution can be defined in terms of step signals and step response. Fig (2.4) shows unit step function and step response. u[n] fig(2.4a) n s[n] fig(2.4b) n 7

34 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. EX 2.5 Find and sketch, the first four few sample value of the impulse response and step responses of the system shown in fig (ex 2.5). Also determine the final value of s[n] as n x(n) y(n) Solution: T By inspection we see that the recurrence formula is (a) Its impulse response is therefore given by Evaluating h[n] term by term we y[n] = 0.8 y[n 1] + x[n] h[n] = 0.8 h[n 1] + δ[n] h[0] = 1, h[1] = 0.8h[0] + δ[1] = 0.8 h[2] = 0.8 h[1] + δ[2] = = 0.64 h[3] = 0.8 h[2] + δ[3] = = (0.8) 3 = and so on as shown in fig below, {s[n] = Σ n h[m]}. (2.4) The step response equals the running sum of h[n].hence its first few value are [using the above results] s[0] = h[0] = 1, s[1] = h[0] + h[1] = = 1.8 s[2] = h[0] + h[1] + h[2] = s[1] + h[2] = = 2.44 s[3] = s[2] + h[3] = (0.8) 3 = s[4] = s[3] + h[4] = and so on. The final value of s[n] forn is given byusing y = 1 + x + x 2 = 1 for x < 1 1 x s[ ] = (0.8) 2 + (0.8) 3 + = = 5 s[n] s[n] n 8 1 n

35 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. EX 2.6 Repeat Ex2.5 find the response to rectangular pulse input with pulse duration n=4 Solution:- We may find the response to the rectangular pulse by consider x[n] as summation of a unit step starting at n=0 and an equal but opposite step starting at n=4. The output is found by superposition two step responses y[n] = s[n] s[n 4] n s[n] s[n-4] y[n] y[n] u[n] n u[n-4] n 1 x[n] n 0 4 n 9

36 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng correlation and convolution Cross-Correlation function is a measure of similarities between two signals. Auto- Correlation is a special case from the cross correlation. Auto correlation represents the cross correlation of the function with itself. Convolution represents the relation between the input and the output of linear system or convolution represents how the input to a system interacts with the system to produce the output. However we clearly need a general computer based a method for doing this a method which will work for any LTI system and any form of input signal. It is known as digital convolution convolution Description The relationship between the input to a linear shift invariant system, x[n] and the output y[n] is given by the convolution sum according the following equation. y[n] = x[n] h[n] = k= x[k]h[n k].. (2.6) Convolution may classify according the sequence:- 1) Circular convolution is used for periodic sequence and is given by N 1 y[n] = x[k] h[n k]. (2.6a) k=0 10

37 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 2) Linear convolution is used for aperiodic sequence and is given by 2N 1 y(n) = k=0 x(k)h(n k). (2.6 b) Where N= sequence length and N = N 1 + N 2 1 wheren 1 =length of x[n] and N 2 =length of h[n] Thus N is the upper limit of the convolution sum. Convolution properties:- a) Commutative:- the commutative property states that the order in which two sequence are convolved is not important. i. e x[n] h[n] = h[n] x[n]. (2.7) From a system point view, this property states that a system with unite impulse response h[n] and input x[n] is the same way as a system with unit sample response x[n] and an input h[n]. b) Associative:- the convolution operator satisfies the associative property {x[n] h 1 [n]} h 2 [n] = x[n] {h 1 [n] h 2 [n]}.. (2.8) Eq(2.8) state that if two systems with impulse response h 1 [n],h 2 [n] are connected in cascade an equivalent system is one that has a unit sample response equal to the convolution of h 1 [n] and h 2 [n]: h eq [n] = h 1 [n] h 2 [n]. (2.9) c) Distributive Property:- the distributive property of the convolution operator state that x[n] {h 1 [n] + h 2 [n]} = x[n] h 1 [n] + x[n] h 2 [n]... (2.10) Eq.(2.8) states that if two with impulse response h 1 [n],h 2 [n] are connected in parallel an equivalent system is one that has a unit sample response equal to the sum h 1 [n] and h 2 [n] h eq [n] = h 1 [n] + h 2 [n]. (2.11) Fig (2.5)shows the interpretation of these properties. 11

38 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Commutation property x[n] y[n] h[n] y[n] h[n] x[n] Associative property x[n] y[n] x[n] y[n] h1[n] h2[n] h1(n) *h2(n) Distributive property x[n] h1[n] y[n] x[n] y[n] + h1(n) +h2(n) h2[n] fig (2.5) the interpretation of convolution properties from a system point view Performing Convolution a) Direct Evaluation: when the sequences that are being convolved may be described by simple closed form mathematical expression, the convolution can be found by direct sum given in eq (2.1). Ex 2.7 Let us perform the convolution of two signals x[n] = a n u[n] = an n 0 0 n < 0 And h[n]=u[n] Solution: u(k) u[k-n] k n u[n-k] k n u[k]u[n-k] k n k 12

39 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. y[n] = x[n] h[n] = x[k]h[n k] k= = a k u[k]u[n k] k= Because u[k]=0 for k<0 and u[n-k] is equal to zero for k>n. when n<0, there are no nonzero terms in the sum and y[n]=0. On the other hand if n 0 n y[n] = a k k=0 Thereforey[n] = 1 an+1 1 a = 1 + a + a 2 + a a n y[n] = 1 an+1 1 a u[n] b) Decomposition of x[n] into a set of weighted shifted impulse method: 1. Plot both x[n] and h[n] as function of n. 2. x[n] is decomposed into a set of weighted, shifted impulses. 3. Generate for each weighted shifted impulses (mentioned in step 2) its own version of the system's impulse response. Each version is weighted by the value of x[n] which cause it, and to begin at the correct instant [this means shift h[n] by the time corresponding each value impulse of x[n]]. 4. The output y[n] is found by superposition of all these individual response. 13

40 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. c) Slide Rule Method Slide rule method is convenient when both x[n] and h[n] are finite in length and short in duration. The steps involved in the slide rule method are as follows:- 1) Write the value of x[k] a long the top of a piece of paper and the value of h[-k] a long the top of another piece of paper as shown in fig (2.5a). 2) Line up the two sequence value x[0] and h[0], multiply each pair of numbers and add the produce to find the value of y[0]. 3) Slide the paper with time reversed sequence h[k] to the right by one, multiply each pair of numbers sum the products to find the value of y[1], and repeat for all shift to the right by n >0. Do the same, shifting the time reversed sequence to the left to find the value of y[n] for n<0. x[-2] x[-1] x[0] x[1] x[2] h[2] h[1] h[0] h[-1] h[-2] Fig (2.5a) the slide rule approaches to convolution. 14

41 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex 2.7Using the decomposition of x[n] into a set of weighted, shifted impulses, to find the output of a digital processor with impulse response h[n] is given by h[n]=0 for n<0 and n>2, h[0]=1, h[1]= -1, h[2]= 2 if the input signal is given by x[n]=0 for n<-1 and n>2, x[0]=2, x[1]=3, x[2]= -1 and x[-1]=1 Solution:- y[n] = k= x[k] h[n k] x[n] 3 h[n] n x[-1]δ(n+1) x[-1]h(n+1) n n n n x[0]δ(n) 2 x[0]h(n) 2 1 n x[1]δ(n-1) x[1]h(n-1) n n x[2]δ(n-2) x[2]δ(n-2) -3 1 n n y[n] n

42 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex 2.8 Find the circular convolution between x[n] = 1, 2, 3,4 n 0 h[n] = 4,3,2,1 n 0 Solution:- the type of the convolution which is used the circular since the number of n is limited [for four value] and this means periodic y[n] = y[n] = N 1 k=0 x[k]h[n k] k=0 3 k=0 x[k]h[n k] h[n] 4 h[n-k] n 2 k=-1 [lag] 3 y[0] = 3 k=0 x[k]h[ k] n y[0] = x[0]h[0] + x[1]h[ 1] + x[2]h[ 2] + x[3]h[ 3] y[0] = = = 24 3 y[1] = x[k]h[1 k] k=0 y[1] = x[0]h[1] + x[1]h[0] + x[2]h[ 1] + x[3]h[ 2] y[1] = = = 22 3 y[2] = x[k]h[2 k] k=0 y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] + x[3]h[ 1] y[2] = = = 24 3 y[3] = x[k]h[3 k] k=0 y[3] = x[0]h[3] + x[1]h[2] + x[2]h[1] + x[3]h[0] y[3] = = 30 h[n-k] y[n] k=1 [lead] 1) Always select The sample of h[n] at n=0 22 n=0 2) h[n-k]=h[-(k-n)]. This is the revere n n 16

43 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex2.9 Find the linear convolution between x[n] = 1,2,3,4 n 0 h[n] = 4,3,2,1 n 0 Solution:- y[n] = 2N 1 k=0 x[k]h[n k] N 1 + N 2 1 = 2N 1 = = 7 since N 1 = N 2 = 4 7 y[n] = x[k]h[n k] k=0 7 y[0] = x[k]h[ k] y[0]= x[0]h[0] + x[1]h[ 1] + x[2]h[ 2] + x[3]h[ 3] + x[4]h[ 4] + x[5]h[ 5] + x[6]h[ 6] + x[7]h[ 7] y[0] = = 4 k=0 Since linear convolution is used for aperiodic sequence and h[n] is only for n 0 h[ 1] = h[ 2] = h[ 3] = h[ 4] = h[ 5] = h[ 6] = h[ 7] = 0 7 y[1] = x[k]h[1 k] k=0 y[1]= x[0]h[1] + x[1]h[0] + x[2]h[ 1] + x[3]h[ 2] + x[4]h[ 3] + x[5]h[ 4] + x[6]h[ 5] + x[7]h[ 6] y[1] = = 11 For the same reason using the same procedure to find y[2]=20, y[3]=30, y[5]=11, y[6]=4, y[7]=0. 7 y[4] = x[k]h[4 k] k=0 y[4]= x[0]h[4] + x[1]h[3] + x[2]h[2] + x[3]h[1] + x[4]h[0] + x[5]h[ 1] + x[6]h[ 2] + x[7]h[ 3] y[4] = = 20.sinceh[n] is exist only for n=0,1,2,3. In this case h[4]=h[5]=h[6]=h[7]=0 17

44 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. The Convolution Integral If the input consists of a continuous of impulses the convolution sum may be replaced by integral and become y(t) = x(τ)h(t τ)dτ y[n] = x[k]h[n k] N n=0 Which is known as the convolution Integral. EX.2.11 Convolve the waveform x(t) and h(t) ((using the convolution integral)) which are defined as follows x(t) = 3 for 0 t 3 h(t) = 2 for 0 t 2 Solution:- 1) Draw x(t) and h(t) as shown in fig (a) 2) The convolution integral depends on the variable τ and so fig (a) has to be replaced by fig(b) 3) Draw the time reverse h(τ) as shown in fig(c) to form h(- τ). 4) h(-τ) is next shift with respect to x(τ) in the direction of positive τ. The resulting waveform h(t-τ) then over laps x(τ) in five geometrical stage.for each of these stages there is corresponding convolution integral. 18

45 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Thus y(t) = x(t) h(t) = + x(τ)y(t τ)dτ x[t] for 0 t 3 Thus x(t) h(t) exists in five stages:- h[t] for 0 t 2 x(t) (a) h(t) x(τ) (b) h(τ) ttττ h(t-τ) 3 x(τ) h(t-τ) 3 x(τ) h(-τ) τττ τ =t 3 (c) (d) stage(1) overlap (e) stage(2a) x(τ) x(τ) x(τ) h(t-τ) overlap h(t-τ) overlap overlap h(t-τ) 0 τ=2 3 0 τ=t-2 τ=t 3 τ=t-2 τ=t τ (f) stage(2b) (g) stage(3) (h) stage(4) 3 x(t) 2 h(t-τ) (i) stage(5) 5 τ (a) Stage 1: t<0 and h(t-τ) does not overlap with x(τ) as shown in fig(d) x(τ) h(t τ) = 0 for all t (b) Stage 2: 0 < t 2 and partial overlap occure between x(τ) and h(t-τ) as shown in fig(e) and (f) y(t) = τ=t 0 x(τ)h(t τ) dτ12 t = 3 2 dτ = 6[τ] t 0 = 6t 0 < t 2 0 The geometrical stage terminates when t=2 as shown in (f) 2 t y(t) 19

46 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. (c) Stage 3: 2 t 3 and there is complete overlap because x(t) and h(t-τ) as shown in fig(g). t y(t) = 3 2 τ=t 2 dτ = [6τ] t t 2 = 12 2 t 3 y(t) (d) Stage 4: 3 t 5 this is another overlap as shown in fig(h) τ=3 y(t) = 3 2 dτ = [6τ] 3 τ=t 2 t 2 12 = 6(5 t) = 30 6t y(t) 2 3 t 3 5 t (e) Stage 5: t>5 as shown in fig(i) there is no overlap y(t) = 0 The convolution integral having a different expression for each of thethreeregion corresponding to three stages as summarized below 0 t 2 y(t)=6t 2 t 3 y(t)=12 3 t 5 y(t)=30-6t y(t) t 20

47 Lect. 2, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 2.6 Difference equation There are many recursive and nonrecursive recurrence formulas to describe the operation of digital processors in the time domain, and are also referred to a difference equations. The general form of a difference equation with three recursive and three nonrecursiveterms is given by y[n] = a 1 y[n 1] + a 2 y[n 2] + a 3 y[n 3] + b 0 x[n 1] + b 1 x[n 1] + b 2 x[n 2] (2.11) To make the equation even more general we allow an arbitrary number of terms and rewrite eq (2.11) as follows a 0 y[n] + a 1 y[n 1] + a 2 y[n 2] + = b 0 x[n] + b 1 x[n 1] + b 2 x[n 2] +.(2.11b) Or N M k=0 a k y[n k] = k=0 b k x[n k] (2.12) The complexity of an LTI processor depends on the number of terms on each side of equation. The value of N which indicates the highest order difference of the output signal is generally referred to as the order of the system. 21

48 University of Technology Department of Electrical Engineering Lect3 Frequency Domain Analysis (Discrete Fourier Series and the Fourier Transform) Asst. Prof. Dr. Hadi T. Ziboon

49 Lect. 3, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 3. Frequency Domain Analysis The Discrete Fourier Series and the Fourier Transform 3.1 Introduction Fourier showed that periodic signals can be represented as weight sums of harmonically related sinusoids. Fourier also, showed that nonrepetitive or aperiodic signal can be considered as the integral of sinusoids which are not harmonically related. These two key ideas form the basis of the famous, Fourier Series and Fourier Transform respectively. Digital computers and special purpose hardware are now used for analyzing the frequency components and the frequency domain performance of systems by using discrete time (digital) techniques. Why do we need also a frequency domain analysis (also we need time domain convolution):- 1) Sinusoidal and exponential signals occur in the natural word, and in the world technology. Even when a signal is not of this type, it can be analyzed into component frequency. 2) If an input signal is described by its frequency spectrum and an LTI system by its frequency response, then the output signal spectrum is found by multiplication both. 3) The design of DSP algorithms and systems often starts with a frequency domain specification. 3.2 Discrete Fourier Series A periodic digital signal can be represented by a Fourier Series and it has a line spectrum similar to analog signal. 1

50 Lect. 3, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Let us consider a periodic digital signal x[n] as shown in fig(3.1), such as x[n] = x[n + N]for all n where N is the period of the sequence x[n]. The coefficients of its line spectrum indicate the amount of various frequencies contained in the signal. They may fined by the following equation C k = 1 N 1 x[n] N n=0 e j2πkn/n. (3.1) Where C k represents the Kth spectral components or harmonic N is the number of number of sample values in each period of the signal Eq(3.1) is known as the analysis equation of the discrete time Fourier series (DTFS). If we know the coefficient C k, we may represent x[n] using the synthesis equation: x[n] = N 1 n=0 C k e j2πkn/n. (3.2) Note: In some texts the1/n multiplier appears in the synthesis equation rather than in the analysis equation and this is not an important difference since it is a scaling factor. one period Re (C k ) Im (C k ) n k k Fig(3.1) (a) Periodic digital signal (b) Real parts of C k (c) Imaginary parts of C k 2

51 Lect. 3, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex(3.1)Determine the spectrum of the signals by using Fourier Series a) x[n] = cos 2πn b) x[n] = 1, 1, 0, 0 n>0 Solution (a) 2πf 0 = 2π f 0 = 1 2 Since f 0 is not a rational number, the signal is not a rational number, the signal is not periodic then the signal cannot expanded in Fourier Series. (b) C k = 1 N N 1 n=0 x[n] = 1 3 x[n] 4 n=0 e 2πjkn 4 3 C k = 1 x[n] 4 n=0 3 e 2πjkn N k=0, 1, 2, 3 e jπkn/2 C 0 = 1 x[n] e 0 = [ ] = 1 2 < 0 n=0 3 C 1 = 1 x[n] 4 n=0 e jπ(1)n 2 = x[n] n=0 e jπn 2 C 1 = 1 4 x[0]e0 + x[1]e jπ 2 + x[2]e jπ + x[3]e j3π 2 C 1 = e jπ = 1 4 [1 + cos π 2 jsin π 2 ] C 1 = 1 2 [1 j] = 4 4 < π 4 In the same way fined C 2 = 0 < 0, C 3 = 2 < π 4 4 C k < C k 1/2 2/4 2/4π/ k k π/4 3

52 Lect. 3, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. Ex(3.2)Sketch the periodic digital signal and find its Fourier CoefficientsC k and sketch their real and imaginary parts x[n] = 1 + sin πn + 2 cos πn 4 4 Solution The values of x[n] are given in the table over one complete period (n=0 to 7) as follows n sin πn 2 cos πn 4 4 x[n] The signal is drawn in fig (a) x[n] can be expressed as a sum of impulses x[n] = δ(n 1) + 0δ(n 2) δ(n 3) + 3δ(n 4) δ(n 5) 2δ(n 6) δ(n 7) The Fourier coefficients are C 3 = 0, C 2 = 1, C 1 = j 2, C 0 = 1, C 1 = j/2, C 2 = 1, C 3 = 0 x[n] n -2 R (C k ) I [C k ] k12 k

53 Lect. 3, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 3.3 Power and Energy of periodic signal Parceval's theorem can be used to calculate the total power or energy of a signal in the time and frequency domains. In the case of a real, periodic digital signal, the theorem takes the form 1 N 1 N n=0 N 1 (3.3) n=0 x[n] 2 = C k 2 The average power of discrete- time periodic signal with period N is defined as: N 1 N 1 n=0 (3.4) P x = 1 N n=0 x[n] 2 = 1 N x[n]x [n] = 1 N 1 x[n]c N n=0 k e j2πkn/n sincex[n] = N 1 n=0 C k e j2πkn/n N 1 = C 1 k N (x[n]e j2πkn N ) n=0 N 1 = C k C k n=0 N 1 = C k 2 n=0 Theenergy of the sequence x[n] over a signal period, then E N = N 1 n=0 x[n] 2 = N N 1 C k 2 n=0 (3.5) 5

54 Lect. 3, Asst. Prof. Dr. Hadi T. Ziboon, Department of Electrical Eng. 3.4 Properties of the Fourier Series In the following discussion we use a double headed arrow to denote the relationship between a signal and its spectrum. Thus x[n] C k indicates that the periodic digital signal x[n] has spectral coefficients C k. x[n] is said to transform into C k. C k inverse transform into x[n]. The main properties of the series:- 1) Linearity If x 1 a k and x 2 b k Then Ax 1 [n] + Bx 2 [n] Aa k + Bb k (3.6) 2) The time shift If x[n] C k Then x[n n 0 ] C k e j2πkn 0/N.. (3.7) 3) The differentiation If x[n] C k Then x[n]-x[n-1] C k [1 e j2πk N ].. (3.8) 4) The integration If x[n] C k and C 0 = 0 n then x[k] k= C k [1 e j2πk/n ] 1 (3.9) 5) Convolution if the digital signals x 1 [n] and x 2 [n] have the same period. x 1 [n] a k and x 2 [n] b k Then N 1 m=0 x 1 [m]x 2 [n m] Na k b k (3.10) The left side of the eq(3.9) expressed a convolution over one period. Eq(3.10) shows that such time domain convolution is equivalent to frequency domain multiplication. 6