SME
Signals & Systems · Problem type
Time scaling and delay in the spectrum
Use this for “x(t) has a part at 5 kHz. Where is it in y(t) = x(2(t − 25 μs)), and what phase does the delay add there?”
Chart
φ(10,000) = −90°
This is an example. Type your own values to start a new problem.
y(t)
=
x(
a
(t
−
t₀ (μs)
))
f₂ (Hz)
=
a
×
f₁ (Hz)
φ
°
=
−360
×
f₂ (Hz)
×
t₀ (μs)
y squeezes x by a, so x’s part at f₁ sits at a × f₁ in y. The delay t₀ then turns that part by −360° × f₂ × t₀.
Formulas
f₂
=
a
×
f₁
10,000 = 2 × 5,000
φ
=
−360
×
f₂
×
t₀
−90
=
−360
×
10,000
×
(2.5
×
10
-5
)
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
Time scaling: x(at) ↔ (1 ÷ a)X(f ÷ a) for a > 0, so squeezing the signal by a moves each part of its spectrum up by a.
•
Delay: x(t − t₀) ↔ X(f)e−j2πft₀, so the magnitude is unchanged and the phase falls by 360° × ft₀.
•
A delay of one whole period, t₀ = 1 ÷ f, adds −360°: the same angle as 0° at that one frequency.
Step-by-step
WE KNOW
a = 2, f₁ = 5,000 Hz, t₀ = 25 μs
FIND
frequency in y (f₂), phase shift (φ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to s).
t₀ = 25 μs = 2.5 × 10⁻⁵ s (1 s = 10⁶ μs)
Step 2 · Find frequency in y (f₂)
Use f₂ = a × f₁
Squeezing
time
by
a
makes
every
part
of
x
repeat
a
times
faster,
so
its
frequency
is
multiplied
by
a
.
f₂
=
2
×
5,000
f₂
=
10,000
Hz
Step 3 · Find phase shift (φ)
Use φ = −360 × f₂ × t₀
The
delay
is
f₂
t₀
periods
of
this
part,
and
each
period
is
360°
of
phase
lost.
φ
=
−360
×
10,000
×
(2.5
×
10
-5
)
φ
=
−3,600,000
×
(2.5
×
10
-5
)
φ
=
−90°
Check
Put the numbers back in. Both sides should match.
10,000 = 2 × 5,000 ✓
−90
=
−360
×
10,000
×
(2.5
×
10
-5
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.