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the problem - agenda (indico)...klystron rf pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the...

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Experimental activity #2: RF pulse compressors Why? β€’ Real devices widely used in electron linac to increase the available peak power β€’ Also very instructive, suitable for an RF lab experience β€’ Based on RF response of resonant cavities and other connection devices (hybrid couplers or circulators) The problem: β€’ High accelerating gradients require very high RF power (∝ 2 ) β€’ RF peak power available from klystrons is typically limited to <100 MW. But the duration of the klystron RF pulses (β‰ˆ few ) may largely exceed the typical filling time of an accelerating structure (<1 )

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Page 1: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Experimental activity #2: RF pulse compressors

Why?

β€’ Real devices widely used in electron linac to increase the available peak power

β€’ Also very instructive, suitable for an RF lab experience

β€’ Based on RF response of resonant cavities and other connection devices (hybrid couplers or

circulators)

The problem:

β€’ High accelerating gradients require very high RF power (𝑃𝑃 ∝ 𝐸𝐸2)

β€’ RF peak power available from klystrons is typically limited to <100 MW. But the duration of the

klystron RF pulses (β‰ˆ few πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating

structure (< 1 πœ‡πœ‡πœ‡πœ‡)

Page 2: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

The Idea: SLED (Stanford Linac Energy Doubler)

β€’ A factor of 4 in power corresponds to a factor of 2 in accelerating gradient, i.e. In beam

energy. Smart idea! But how a large RF power flowing in a waveguide can be

manipulated in order to be compressed?

β€’ This idea has been firstly implemented at SLAC in the 70’s [1]. The compression is

obtained by capturing the RF power reflected by high-Q resonant cavities.

4 πœ‡πœ‡πœ‡πœ‡

1 πœ‡πœ‡πœ‡πœ‡

200 𝑀𝑀𝑀𝑀50 𝑀𝑀𝑀𝑀Pulse Compression

π‘ƒπ‘ƒπ‘ƒπ‘ƒπ‘ƒπ‘ƒπœ‡πœ‡π‘ƒπ‘ƒ 𝐸𝐸𝐸𝐸𝑃𝑃𝐸𝐸𝐸𝐸𝐸𝐸 β‰ˆ 200 π½π½π‘˜π‘˜π‘ƒπ‘ƒπΈπΈπœ‡πœ‡π‘˜π‘˜πΈπΈπ‘˜π‘˜πΈπΈ 𝑅𝑅𝑅𝑅 π‘π‘π‘ƒπ‘ƒπ‘ƒπ‘ƒπœ‡πœ‡π‘ƒπ‘ƒ

β€’ Let’s compress the energy of the RF pulse in about 1 filling time to increase the peak

power to the maximum!

[1] Farkas et al. - SLED A method of doubling SLAC energy

Page 3: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

walls

out

extoutext P

PQQ

PUQ === 0; Ξ²Ο‰

The extra-power that flows through the cavity couplers may significantly change the characteristics of theresonance. This effect is known as β€œcavity loading”. The loaded cavity Q-factor is lowered by the powercoupled out through the cavity ports and results to be:

+=

+=

β†’+=β†’+

==βˆ‘

βˆ‘βˆ‘βˆ‘

0

0

11

111

1

QQ

QQQ

UP

UP

QPPU

PUQ

n

L

extLoutwalls

LoutwallsTotL

nn

n βωωωω

where 𝑄𝑄0 is the usual quality factor of the resonant mode, related only to the dissipation on the cavity walls

Coupling coefficients of a resonant cavity

β€’ Β«RealΒ» resonant cavities are NOT perfectly closed volumes,i.e. at least 1 opening (port) must be present to allow the RFpower to flow in (and out)

β€’ The coupling strength of a port can be measured as theamount of power π‘ƒπ‘ƒπ‘œπ‘œπ‘œπ‘œπ‘œπ‘œ extracted from the cavity throughthe port itself for a given level of the mode fields inside

β€’ This leads to the definitions of the external-Q (π‘„π‘„π‘’π‘’π‘’π‘’π‘œπ‘œ) (inanalogy with the definition of the resonance quality factor𝑄𝑄0) and coupling coefficient of a coupler according to :

Page 4: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Wave reflection in guide terminated on a resonant load

𝑍𝑍𝑐𝑐𝑐𝑐𝑐𝑐 πœ‡πœ‡ = 𝛽𝛽 𝑍𝑍0β„πœ‡πœ‡ (πœ”πœ”0𝑄𝑄0)

β„πœ‡πœ‡ πœ”πœ”02 + β„πœ‡πœ‡ (πœ”πœ”0𝑄𝑄0) + 1

𝛽𝛽 = input coupling coefficient

πœ”πœ”02 =

1𝐿𝐿𝐿𝐿

, 𝑄𝑄0 = πœ”πœ”0𝛽𝛽𝑍𝑍0𝐿𝐿

β€’ A resonant cavity is a non-matched load for a waveguide or a transmission line, and can

be modeled as an RLC parallel circuit.

β€’ An RF wave travelling from the generator toward the cavity is always partially reflected at

the cavity input. The reflected wave can be calculated according to:

𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑉𝑉𝑅𝑅𝐹𝐹𝐹𝐹𝑍𝑍𝑐𝑐𝑐𝑐𝑐𝑐 βˆ’ 𝑍𝑍0𝑍𝑍𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑍𝑍0

Page 5: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Cavity reflected waves and input coupling coefficient

𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑉𝑉𝑅𝑅𝐹𝐹𝐹𝐹𝑍𝑍𝑐𝑐𝑐𝑐𝑐𝑐 βˆ’ 𝑍𝑍0𝑍𝑍𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑍𝑍0

= 𝑉𝑉𝑅𝑅𝐹𝐹𝐹𝐹2𝛽𝛽𝛽𝛽 + 1

β„πœ‡πœ‡ πœ”πœ”0π‘„π‘„πΏπΏβ„πœ‡πœ‡ πœ”πœ”0

2 + β„πœ‡πœ‡ πœ”πœ”0𝑄𝑄𝐿𝐿 + 1βˆ’ 1

Lorentzian network transfer function

β€’ Depending on the value of 𝛽𝛽, 3 practical situations can occur:

𝛽𝛽 < 1 undercoupling

𝛽𝛽 = 1 critical coupling

𝛽𝛽 > 1 overcoupling

1. Calculate and plot the Lorentzian network response when VIN :

2. Calculate and plot the cavity reflected wave when VFWD :

β€’ In the previous equation, if we substitute the expression for 𝑍𝑍𝑐𝑐𝑐𝑐𝑐𝑐, the reflected wave becomes function of the

coupling coefficient:

Step pulse

Rectangular pulse

Step pulse

Rectangular pulse

What are we going to do:

Page 6: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Lorentzian network response: RF step pulse

π‘½π‘½π’Šπ’Šπ’Šπ’Š(𝒕𝒕) = π‘½π‘½πŸŽπŸŽ οΏ½ 𝟏𝟏(𝒕𝒕 βˆ’ 𝑻𝑻𝒔𝒔𝒕𝒕𝒔𝒔𝒔𝒔𝒕𝒕) οΏ½ 𝒄𝒄𝒄𝒄𝒔𝒔 𝝎𝝎𝟎𝟎(𝒕𝒕 βˆ’ 𝑻𝑻𝒔𝒔𝒕𝒕𝒔𝒔𝒔𝒔𝒕𝒕 )

𝑽𝑽𝒄𝒄𝒐𝒐𝒕𝒕 𝒕𝒕 = π‘½π‘½πŸŽπŸŽ οΏ½ 𝟏𝟏 𝒕𝒕 βˆ’ 𝑻𝑻𝒔𝒔𝒕𝒕𝒔𝒔𝒔𝒔𝒕𝒕 οΏ½ 𝒄𝒄𝒄𝒄𝒔𝒔 𝝎𝝎𝟎𝟎(π’•π’•βˆ’π‘»π‘»π’”π’”π’•π’•π’”π’”π’”π’”π’•π’•) οΏ½ 𝟏𝟏 βˆ’ π’†π’†βˆ’ ⁄(π’•π’•βˆ’π‘»π‘»π’”π’”π’•π’•π’”π’”π’”π’”π’•π’•) 𝝉𝝉 ; with 𝝉𝝉 = β„πŸπŸπ‘Έπ‘Έπ‘³π‘³ 𝝎𝝎𝟎𝟎

𝑄𝑄𝐿𝐿 = 25000;πœ”πœ”0 = 2856 𝑀𝑀𝑀𝑀𝑀𝑀;π‘‡π‘‡π‘ π‘ π‘œπ‘œπ‘π‘π‘ π‘ π‘œπ‘œ = 0.5 πœ‡πœ‡πœ‡πœ‡; 𝜏𝜏 β‰ˆ 2. 8πœ‡πœ‡πœ‡πœ‡

π‘‡π‘‡π‘ π‘ π‘œπ‘œπ‘π‘π‘ π‘ π‘œπ‘œ

The response of a Lorentzian network to an RF step pulse is:

Page 7: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

𝑄𝑄𝐿𝐿 = 25000;πœ”πœ”0 = 2856 𝑀𝑀𝑀𝑀𝑀𝑀;π‘‡π‘‡π‘ π‘ π‘œπ‘œπ‘π‘π‘ π‘ π‘œπ‘œ = 0.5 πœ‡πœ‡πœ‡πœ‡;𝑇𝑇𝑒𝑒𝑒𝑒𝑒𝑒 = 5.5 πœ‡πœ‡πœ‡πœ‡; 𝜏𝜏 β‰ˆ 2. 8πœ‡πœ‡πœ‡πœ‡

π‘‡π‘‡π‘’π‘’π‘’π‘’π‘’π‘’π‘‡π‘‡π‘ π‘ π‘œπ‘œπ‘π‘π‘ π‘ π‘œπ‘œ

Lorentzian network response: RF rectangular pulse

π‘½π‘½π’Šπ’Šπ’Šπ’Š(𝒕𝒕) = π‘½π‘½πŸŽπŸŽ οΏ½ 𝒄𝒄𝒄𝒄𝒔𝒔 𝝎𝝎𝟎𝟎(𝒕𝒕 βˆ’ 𝑻𝑻𝒔𝒔𝒕𝒕𝒔𝒔𝒔𝒔𝒕𝒕) οΏ½ 𝟏𝟏 𝒕𝒕 βˆ’ 𝑻𝑻𝒔𝒔𝒕𝒕𝒔𝒔𝒔𝒔𝒕𝒕 βˆ’ 𝟏𝟏(𝒕𝒕 βˆ’ π‘»π‘»π’†π’†π’Šπ’Šπ’†π’†)

𝑽𝑽𝒄𝒄𝒐𝒐𝒕𝒕 𝒕𝒕 = π‘½π‘½πŸŽπŸŽ οΏ½ 𝒄𝒄𝒄𝒄𝒔𝒔 𝝎𝝎𝟎𝟎(𝒕𝒕 βˆ’ 𝑻𝑻𝒔𝒔𝒕𝒕𝒔𝒔𝒔𝒔𝒕𝒕) �𝟏𝟏 𝒕𝒕 βˆ’ 𝑻𝑻𝒔𝒔𝒕𝒕𝒔𝒔𝒔𝒔𝒕𝒕 οΏ½ 𝟏𝟏 βˆ’ π’†π’†βˆ’ β„π’•π’•βˆ’π‘»π‘»π’”π’”π’•π’•π’”π’”π’”π’”π’•π’• 𝝉𝝉 +βˆ’πŸπŸ 𝒕𝒕 βˆ’ π‘»π‘»π’†π’†π’Šπ’Šπ’†π’† οΏ½ 𝟏𝟏 βˆ’ π’†π’†βˆ’( β„π’•π’•βˆ’π‘»π‘»π’†π’†π’Šπ’Šπ’†π’†) 𝝉𝝉 with 𝝉𝝉 = β„πŸπŸπ‘Έπ‘Έπ‘³π‘³ 𝝎𝝎𝟎𝟎

The response of a Lorentzian network to an RF rectangular pulse is:

Page 8: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

RF step pulse reflected by a resonant cavity

𝑉𝑉𝑅𝑅𝐹𝐹𝐹𝐹(π‘˜π‘˜) = 𝑉𝑉0 οΏ½ 1(π‘˜π‘˜) οΏ½ π‘π‘π‘˜π‘˜πœ‡πœ‡ πœ”πœ”0π‘˜π‘˜ ; 𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 π‘˜π‘˜ = 𝑉𝑉0 οΏ½ 1 π‘˜π‘˜ οΏ½ π‘π‘π‘˜π‘˜πœ‡πœ‡ πœ”πœ”0π‘˜π‘˜ οΏ½2𝛽𝛽𝛽𝛽 + 1 1 βˆ’ π‘ƒπ‘ƒβˆ’ β„π‘œπ‘œ 𝜏𝜏 βˆ’ 1

Undercoupled, 𝛽𝛽 < 1

Critically coupled, 𝛽𝛽 = 1

Overcoupled, 𝛽𝛽 > 1

𝑉𝑉𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑉𝑉𝑅𝑅𝐹𝐹𝐹𝐹2𝛽𝛽𝛽𝛽 + 1

β„πœ‡πœ‡ πœ”πœ”0π‘„π‘„πΏπΏβ„πœ‡πœ‡ πœ”πœ”0

2 + β„πœ‡πœ‡ πœ”πœ”0𝑄𝑄𝐿𝐿 + 1βˆ’ 1

In time domain:

The reflected wave of a resonant cavity to an RF step pulse has the following transfer function:

Page 9: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

RF rectangular pulse reflected by an over-coupled resonant cavity

𝑉𝑉𝑅𝑅𝐹𝐹𝐹𝐹(π‘˜π‘˜) = 𝑉𝑉0 οΏ½ π‘π‘π‘˜π‘˜πœ‡πœ‡ πœ”πœ”0π‘˜π‘˜ οΏ½ 1 π‘˜π‘˜ βˆ’ 1(π‘˜π‘˜ βˆ’ 𝑇𝑇𝑝𝑝)

𝑉𝑉𝑅𝑅𝑅𝑅𝐿𝐿 π‘˜π‘˜ = 𝑉𝑉0 οΏ½ π‘π‘π‘˜π‘˜πœ‡πœ‡ πœ”πœ”0π‘˜π‘˜ οΏ½ 1 π‘˜π‘˜2𝛽𝛽𝛽𝛽 + 1 1 βˆ’ π‘ƒπ‘ƒβˆ’ β„π‘œπ‘œ 𝜏𝜏 βˆ’ 1 βˆ’ 1 π‘˜π‘˜ βˆ’ 𝑇𝑇𝑝𝑝

2𝛽𝛽𝛽𝛽 + 1 1 βˆ’ π‘ƒπ‘ƒβˆ’ β„π‘œπ‘œβˆ’π‘‡π‘‡π‘π‘ 𝜏𝜏 βˆ’ 1

The RFL peak field is maximum at the end of theFWD pulse, and its value exceeds the peak of theFWD itself.

This effect can be enhanced by forcing a 180Β° phasejump in the driving pulse at a certain time before theend of the pulse. By doing that we induce anequivalent step in the driving signal of doubleamplitude which adds in phase with the signaloriginated by the first rising front of the driving RFpulse.

𝑇𝑇𝑝𝑝

𝑇𝑇𝑠𝑠

𝑉𝑉𝑅𝑅𝑅𝑅𝐿𝐿 π‘˜π‘˜ = 𝑉𝑉0 οΏ½ π‘π‘π‘˜π‘˜πœ‡πœ‡ πœ”πœ”0π‘˜π‘˜ οΏ½ 1 π‘˜π‘˜2𝛽𝛽𝛽𝛽 + 1

1 βˆ’ π‘ƒπ‘ƒβˆ’ β„π‘œπ‘œ 𝜏𝜏 βˆ’ 1 βˆ’ 2 οΏ½ 1 π‘˜π‘˜ βˆ’ 𝑇𝑇𝑠𝑠2𝛽𝛽𝛽𝛽 + 1

1 βˆ’ π‘ƒπ‘ƒβˆ’ β„π‘œπ‘œβˆ’π‘‡π‘‡π‘ π‘  𝜏𝜏 βˆ’ 1 + 1 π‘˜π‘˜ βˆ’ 𝑇𝑇𝑝𝑝2𝛽𝛽𝛽𝛽 + 1

1 βˆ’ π‘ƒπ‘ƒβˆ’ β„π‘œπ‘œβˆ’π‘‡π‘‡π‘π‘ 𝜏𝜏 βˆ’ 1

𝑉𝑉𝑅𝑅𝐹𝐹𝐹𝐹(π‘˜π‘˜) = 𝑉𝑉0 οΏ½ π‘π‘π‘˜π‘˜πœ‡πœ‡ πœ”πœ”0π‘˜π‘˜ οΏ½ 1 π‘˜π‘˜ βˆ’ 2 οΏ½ 1 π‘˜π‘˜ βˆ’ 𝑇𝑇𝑠𝑠 + 1(π‘˜π‘˜ βˆ’ 𝑇𝑇𝑝𝑝)

The wave reflected by an over-coupled cavity excited by a rectangular RF pulse is:

Page 10: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

βˆ’=

001

001

100

100

21

j

j

j

j

S

=

010001100

S

Circulators are non-reciprocal (typically) 3-portsdevices whose scattering matrix is ideally given by:

How to route the power flow towards the accelerator

(and not towards the power units) 90Β° hybrids are non-reciprocal (typically) 4-ports deviceswhose scattering matrix is ideally given by:

Page 11: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Experimental activity #2

1. Introduction: β€’ Measure environmental conditions and calculate SLED tuning

frequency (working temperature Tr = 35 Β°C, and nominal frequency fRF = 2856 MHz)

2. Time domain: set-up of the RF pulseβ€’ CW signal at fLP from signal generatorβ€’ Stanford setup (5 us, 100 Hz, TTL HighZ)β€’ RF Switch setupβ€’ Oscilloscope cross-check

3. Frequency domain: realization of the 180Β° phase jumpβ€’ Design of a 180Β° phase shifting circuitβ€’ Functionality cross-check of the 180Β° phase shifter providedβ€’ Actual phase shift measurementβ€’ Generation and timing regulation of the trigger signalsβ€’ Oscilloscope cross check (30 dB att. 500 mV/div)

4. Time domain: SLED tuning with rectangular input pulse (no 180Β° phase jump)

1. Cavity n.1 optimization 2. Cavity n.2 optimization3. Tuning optimizing SLED output

5. Time domain: visualization of SLED output pulse with 180Β°phase jump

1. Fine frequency tuning of the SLED pulse2. Qualitative measurement of energy/power gain3. Scan Tjump vs energy gain

Input port

90Β° hybrid

High Q cavities

Output port

Page 12: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Hint: SLED tuning

WHY?If the resonant frequencies of the 2 cavities are not the same and/or equal to the operatingfrequency (2856 MHz for S-band linacs) the SLED can produce high reflections at the inputport, causing breakdowns in the input waveguide or near the klystron window.

HOW?Generally frequency tuning is done in air and with low RF power. For this reason themeasurement needs to account for several environmental variation with respect to nominalconditions: room temperature, pressure, humidity and air (instead of vacuum).

𝑓𝑓𝐿𝐿𝐿𝐿 𝑀𝑀𝑀𝑀𝑀𝑀 = 𝑓𝑓𝐻𝐻𝐿𝐿 𝑀𝑀𝑀𝑀𝑀𝑀 + βˆ†π‘“π‘“π‘‡π‘‡(𝑀𝑀𝑀𝑀𝑀𝑀) βˆ’ βˆ†π‘“π‘“π‘π‘(𝑀𝑀𝑀𝑀𝑀𝑀)

Where:𝑓𝑓𝐻𝐻𝐿𝐿 is the SLED nominal working frequency = 2856 MHz𝑓𝑓𝐿𝐿𝐿𝐿 is the SLED resonant frequency to be tuned at low power, room temperature and in airβˆ†π‘“π‘“π‘‡π‘‡ is the frequency correction for temperatureβˆ†π‘“π‘“π‘π‘ is the frequency correction for pressure and humidity

Page 13: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Hint: SLED tuning (finding fLP)

EMPIRICAL FORMULAE (from SLAC & IHEP experience)

βˆ†π‘“π‘“π‘‡π‘‡= 0.045 𝑇𝑇𝑠𝑠 βˆ’ π‘‡π‘‡π‘šπ‘š 𝑀𝑀𝑀𝑀𝑀𝑀 β†’ 45 π‘˜π‘˜π‘€π‘€π‘€π‘€/°𝐿𝐿

βˆ†π‘“π‘“π‘π‘=0.2856

273 + π‘‡π‘‡π‘šπ‘š798 + 𝑝𝑝𝑀𝑀 0.9 1 +

5580π‘‡π‘‡π‘šπ‘š

βˆ’ 1.05 𝑀𝑀𝑀𝑀𝑀𝑀

𝑝𝑝𝑀𝑀 =𝑅𝑅

1004.58 + 0.092 οΏ½ π‘‡π‘‡π‘šπ‘š1.65 𝑀𝑀𝑀𝑀𝑀𝑀

Where:

𝑇𝑇𝑠𝑠 is the operating temperature (in Β°C) in high power;π‘‡π‘‡π‘šπ‘š is the room temperature (in Β°C) during low power measurement;𝑅𝑅 is the relative humidity (in %) during low power measurement.

Page 14: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Hint: SLED tuning (single cavities)

β€’ Connect the output waveguide to a matched load;β€’ Set-up the RF pulse:

1. CW signal at fLP

2. Pulsed signal at fLP (5 us length, 100 Hz rep. rate)

β€’ Display the reflection signal from each cavity on an oscilloscope (use a flat pulse without phase reversal). The connections should look like the scheme below:

β€’ Rotate the tuning screw to tune the resonant frequency of the cavity to have a minimum (dashed line). Repeat the procedure a couple of times for fine tuning.

β€’ Check the input VSWR: it should not be higher than 1.2

OSCILLOSCOPE

Page 15: The problem - Agenda (Indico)...klystron RF pulses (β‰ˆfew πœ‡πœ‡πœ‡πœ‡) may largely exceed the typical filling time of an accelerating structure (< 1 πœ‡πœ‡πœ‡πœ‡) ... This

Hint: SLED tuning (sum of 2 cavities)

β€’ Once the cavities are individually tuned, the output signal of the SLED (without phase jump) should look like the one below:

β€’ Rotate the tuning screw to fine-tune the SLED output

β€’ If the cavities do not have the same response (different Q0 or different Ξ²), the optimum might not be where one expects…

OSCILLOSCOPE


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