Overview of Sampling Topics • (Shannon) sampling theorem • Impulse-train sampling • Interpolation (continuous-time signal reconstruction) • Aliasing • Relationship of CTFT to DTFT • DT processing of CT signals • DT sampling • Decimation & interpolation J. McNames Portland State University ECE 223 Sampling Ver. 1.15 1
Amplitude versus Time • In this class we are working only with continuous-valued signals • Signals with discrete values that have been quantized are called digital signals • Analog-to-Digital converters (ADC) convert continuous-valued signals to discrete-valued signals. These often also convert from CT to DT. • Digital-to-analog converters (DAC) convert discrete-valued signals to continuous-valued signals. These often also convert from DT to CT. • Signals that are both continuous-valued and continuous-time are usually called analog signals • Signals that are both discrete-valued and discrete-time are usually called digital J. McNames Portland State University ECE 223 Sampling Ver. 1.15 2
Overview of DT Processing of CT Signals x [ n ] y [ n ] H ( z ) x ( t ) y ( t ) CT ⇒ DT DT ⇒ CT T s T s • Many systems (1) sample a signal, (2) process it in discrete-time, and (3) convert it back to a continuous-time signal • Called discrete-time processing of continuous-time signals • Most modern digital signal processing (DSP) uses this architecture • The first step is called sampling • The last step is called interpolation J. McNames Portland State University ECE 223 Sampling Ver. 1.15 3
Signals x ( t ) → x [ n ] → x i ( t ) X ( jω ) → X (e jω ) → X i ( jω ) • For now, just consider the two conversions – Sampling (CT → DT) – Interpolation (DT → CT) • Suppose we want x i ( t ) to be as close to x ( t ) as possible • Need to consider relationships in both time- and frequency-domains • Three signals, three transforms J. McNames Portland State University ECE 223 Sampling Ver. 1.15 4
Guiding Questions and Objectives x [ n ] = x ( t ) | t = nT s • The operation of sampling (CT → DT conversion) is trivial • The challenge is to understand the limits and tradeoffs • The remainder of these slides is dedicated to answering the following three questions 1. How is the CTFT of x ( t ) , X ( jω ) related to the DTFT of x [ n ] = x ( nT s ) , X (e j Ω ) ? 2. Under what conditions can we synthesize x ( t ) from x [ n ] : x [ n ] → x ( t ) ? 3. How do we perform this DT → CT conversion? J. McNames Portland State University ECE 223 Sampling Ver. 1.15 5
Need for a Bridge Signal Consider three signals Type Time Domain Frequency Domain Original Signal CT x ( t ) X ( jω ) X (e j Ω ) Sampled Signal DT x [ n ] Bridge Signal CT x δ ( t ) X δ ( jω ) • Goals: to determine the relationship of x ( t ) to x [ n ] = x ( nT s ) in the time and frequency domains • In the time domain, the relationship x [ n ] = x ( nT s ) is clear • But what is the relationship of X ( jω ) to X (e j Ω ) ? • The transforms differ in character – X (e j Ω ) is periodic – X (e j Ω ) has units of radians per sample • A bridge signal x δ ( t ) is the only way (that I know of) to determine how X ( jω ) is related to X (e j Ω ) J. McNames Portland State University ECE 223 Sampling Ver. 1.15 6
Use of the Bridge Signal • Define the bridge signal by relating X δ ( jω ) to X (e j Ω ) • Determine how x ( t ) is related to x δ ( t ) in the time domain • Determine how X ( jω ) is related to X δ ( jω ) • Use the relationships of X (e j Ω ) and X ( jω ) to X δ ( jω ) to determine their relationship to one another J. McNames Portland State University ECE 223 Sampling Ver. 1.15 7
Defining the Bridge Signal • The bridge signal x δ ( t ) is a CT representation of a DT signal x [ n ] • It is defined as having the same transform, within a scale factor, as the DT signal • Suppose x [ n ] = A e j (Ω n + φ ) • What is the CT equivalent? • Let us pick x ( t ) = A e j ( ωt + φ ) • How do we relate the DT frequency Ω to the CT frequency ω ? – ω has units of radians/second – Ω has units of radians/sample • Let us use the conversion factor of T s = f − 1 seconds/sample s Ω (radians/sample) = ω (radians/second) × T s (seconds/sample) ω (radians/second) = Ω (radians/sample) × f s (samples/second) J. McNames Portland State University ECE 223 Sampling Ver. 1.15 8
Defining the Bridge Signal X δ ( jω ) = X (e j Ω ) � Ω= ωT s = X (e jωT s ) � • The bridge signal is a CT representation of a DT signal • We define it by equating the CTFT and DTFT with an appropriate scaling factor for frequency • Note that this is a highly unusual CT signal – The CTFT is periodic – What does this tell us about x δ ( t ) ? J. McNames Portland State University ECE 223 Sampling Ver. 1.15 9
Solving for the Bridge Signal ∞ � X (e j Ω ) = x [ n ] e − j Ω n n = −∞ ∞ � X δ ( jω ) = X (e j Ω ) � x [ n ] e − jωT s n Ω= ωT s = � n = −∞ FT x δ ( t ) ⇐ ⇒ X δ ( jω ) FT ⇒ e − jωt o δ ( t − t o ) ⇐ FT ⇒ e − jωT s n δ ( t − T s n ) ⇐ ∞ ∞ FT � � x [ n ] e − jωT s n x [ n ] δ ( t − T s n ) ⇐ ⇒ n = −∞ n = −∞ ∞ � x δ ( t ) = x [ n ] δ ( t − T s n ) n = −∞ J. McNames Portland State University ECE 223 Sampling Ver. 1.15 10
Why Isn’t the Inverse Transform Similar to x ( t ) ? x [ n ] = 1 � X (e jω ) e j Ω n dΩ DTFT 2 π 2 π � + ∞ x ( t ) = 1 X ( jω ) e jωt d ω CTFT 2 π −∞ • We generated x δ ( t ) from x [ n ] by equating a CTFT to the DTFT of x [ n ] • This seems reasonable at first • But consider the range of the CTFT and DTFT synthesis equations • The DTFT synthesizes x [ n ] out of a finite range of frequencies • The CTFT synthesizes x ( t ) out of all frequencies • This is why, if x [ n ] = x ( nT s ) , that x δ ( t ) = F − 1 { X δ ( jω ) } = F − 1 � X (e jωT s ) � � = x ( t ) J. McNames Portland State University ECE 223 Sampling Ver. 1.15 11
Impulse Sampling p ( t ) 1 -4Ts -3Ts -2Ts -Ts 0 Ts 2Ts 3Ts 4Ts 5Ts ∞ � p ( t ) = δ ( t − nT s ) n = −∞ ∞ � x δ ( t ) = x ( nT s ) δ ( t − T s n ) = x ( t ) p ( t ) n = −∞ • x δ ( t ) can also be formed from a CT signal • This is called impulse sampling • We can model sampling by use of the periodic impulse train J. McNames Portland State University ECE 223 Sampling Ver. 1.15 12
Rectangular Window and Impulse Train Notation p ( t ) 1 -4Ts -3Ts -2Ts -Ts 0 Ts 2Ts 3Ts 4Ts 5Ts • Note that a similar symbol was used for rectangular windows � 1 | t | < T p T ( t ) = 0 Otherwise but p ( t ) � = p T ( t ) J. McNames Portland State University ECE 223 Sampling Ver. 1.15 13
Impulse Sampling Conceptual Example x ( t ) 1 -4Ts -3Ts -2Ts -Ts 0 Ts 2Ts 3Ts 4Ts 5Ts p ( t ) 1 -4Ts -3Ts -2Ts -Ts 0 Ts 2Ts 3Ts 4Ts 5Ts x ( t ) p ( t ) 1 -4Ts -3Ts -2Ts -Ts 0 Ts 2Ts 3Ts 4Ts 5Ts J. McNames Portland State University ECE 223 Sampling Ver. 1.15 14
Impulse Sampling Terminology x [ n ] y [ n ] H ( z ) x ( t ) y ( t ) CT ⇒ DT DT ⇒ CT T s T s ∞ � x δ ( t ) = x ( t ) p ( t ) = x ( nT s ) δ ( t − nT s ) n = −∞ • The impulse train p ( t ) is called the sampling function • p ( t ) is periodic with fundamental period T s • T s , the fundamental period of p ( t ) , is called the sampling period T s and ω s = 2 π 1 • f s ≡ T s are called the sampling frequency J. McNames Portland State University ECE 223 Sampling Ver. 1.15 15
Fourier Transforms of Periodic Signals Overview ∞ � x δ ( t ) = x ( nT s ) δ ( t − T s n ) = x ( t ) p ( t ) n = −∞ X δ ( jω ) = 1 2 π X ( jω ) ∗ P ( jω ) • To determine how X δ ( jω ) is related to X ( jω ) , we need to calculate P ( jω ) • p ( t ) is a periodic function with infinite energy • The CTFT clearly doesn’t converge • Is easier to – Calculate the Fourier series coefficients P [ k ] for p ( t ) – Solve for P ( jω ) from P [ k ] using the general relationship between the CTFS and the CTFT • Since periodic signals have infinite energy, the CTFT of these signals consists of impulses J. McNames Portland State University ECE 223 Sampling Ver. 1.15 16
Fourier Transforms of Periodic Signals Recall the Fourier series representations of periodic signals ∞ � X [ k ] e jkωt x ( t ) = k = −∞ FT e jω o t ⇐ ⇒ 2 π δ ( ω − ω o ) ∞ ∞ FT � � X [ k ] e jkω o t x ( t ) = ⇐ ⇒ 2 π X [ k ] δ ( ω − kω o ) k = −∞ k = −∞ J. McNames Portland State University ECE 223 Sampling Ver. 1.15 17
Example 1: CT Fourier Transform of an Impulse Train Solve for the Fourier transform of the impulse train ∞ � p ( t ) = δ ( t − nT s ) n = −∞ Hint: the impulse train is periodic. � + ∞ X [ k ] = 1 � x ( t )e − jkω o t d t x ( t ) e − jωt d t X ( jω ) = T T −∞ J. McNames Portland State University ECE 223 Sampling Ver. 1.15 18
Example 1: Workspace J. McNames Portland State University ECE 223 Sampling Ver. 1.15 19
Recommend
More recommend
